450 + Interview Questions in LIMITS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	`lim_(xto oo) (1)/(n) {1+e^(1//n)+e^(2//n)+e^(3//n)+.....+e^((n-1)/(n))}` is equal toA. eB. `-e`C. `e-1`D. `1-e`
Answer» Correct Answer - C We have, `lim_(xto oo) (1)/(n){1+e^(1//n)+e^(2//n)+.....+e^((n-1)/(n))}` ` =lim_(xto oo) (1)/(n){(1-(e^(1//n))^n)/(1-e^(1//n))}=lim_(xto oo) (1-e)/(((1-e^(1//n))/(1//n)))=(1-e)/(-1)=e-1`.

Discussion

2.	`lim_(xto oo) ((nsqrt(a)+nsqrt(b))/(2))^n,a,b,gt 0` equalsA. 1B. `sqrt(pq)`C. `pq`D. `(pq)/(2)`
Answer» Correct Answer - B `lim_(ntooo) ((nsqrt(p)+nsqrt(q))/(2))` ` =lim_(xto0) ((p^x+q^x)/(2))^(1//x)`, where `n=(1)/(x)` ` lim_(xto0) [1+(1)/(2) {(p^x-1)+(q^x-1)}]^(1//x)=sqrt(pq)`

Discussion

3.	If `lim_(xto oo){(x^2+1)/(x+1)-(ax+b)}to oo`, thenA. `a in (1,oo)`B. `a ne 1 , b in R`C. `a in (-oo,1)`D. none of these
Answer» Correct Answer - C We have, `lim_(n to oo) {(x^2+1)/(x+1)-ax-b}=oo` `rArr lim_(x to oo) (x^2(1-a)-x(a+b)+1-b)/(x+1)=oo` The limit of the given expression will be infinity, if degree of numberator is more than that of denominaotor. `therefore 1-a gt 0rArr alt 1`. Hence, `a in (- oo,1)` can assume any real value.

Discussion

4.	Evaluate:\(\lim\limits_{x \to 1}\frac{\sqrt {1+x}-\sqrt {1-x}}{1+x}\)
Answer» \(\lim\limits_{x \to 1}\frac{\sqrt {1+x}-\sqrt {1-x}}{1+x}\) \(=\lim\limits_{x \to 1}\frac{\sqrt {1+1}-\sqrt {1-1}}{1+1}\) \(=\frac{\sqrt 2-0}{2}\) \(=\frac{\sqrt 2}{2}\) \(=\frac{1}{\sqrt 2}\)

4.

Evaluate:\(\lim\limits_{x \to 1}\frac{\sqrt {1+x}-\sqrt {1-x}}{1+x}\)

Answer»

\(\lim\limits_{x \to 1}\frac{\sqrt {1+x}-\sqrt {1-x}}{1+x}\)

\(=\lim\limits_{x \to 1}\frac{\sqrt {1+1}-\sqrt {1-1}}{1+1}\)

\(=\frac{\sqrt 2-0}{2}\)

\(=\frac{\sqrt 2}{2}\)

\(=\frac{1}{\sqrt 2}\)

Discussion

5.	\(f\left(x\right)=\left[\left(\frac{min\left(t^2+4t+6\right)\sin \left(x\right)}{x}\right)\right]\)where [.] represents greatest integer function, then find \(\lim _{x\to 0}\left(f\left(x\right)\right)\)
Answer» f(x) = \([\frac{min(t^2+4t+6)sin x}x]\) ∵ t² + 4t + 6 = (t + 2)² + 2 \(\geq\) 2 ∴ min (t² + 4t + 6) = 2 ∴ f(x) = \([\frac{2sinx}x]\) ∵ sin x \(\leq\) x ⇒ -1 < \(\frac{sin x}x\leq1\) ⇒ -2 < \(\frac{2sin x}x\leq2\) ∴ \([\frac{2sinx}x]\) = \(\begin{cases}2&;x=0or sinx/x=1\\1&;1/2 \leq\frac{sin x}x<1\\0&;0\leq\frac{sin x}x<1/2\\-1&;\frac{sin x}x<0\end{cases}\) \(\lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}[\frac{2sin x}x]\) = 1

5.

\(f\left(x\right)=\left[\left(\frac{min\left(t^2+4t+6\right)\sin \left(x\right)}{x}\right)\right]\)where [.] represents greatest integer function, then find \(\lim _{x\to 0}\left(f\left(x\right)\right)\)

Answer»

f(x) = \([\frac{min(t^2+4t+6)sin x}x]\)

∵ t² + 4t + 6 = (t + 2)² + 2 \(\geq\) 2

∴ min (t² + 4t + 6) = 2

∴ f(x) = \([\frac{2sinx}x]\)

∵ sin x \(\leq\) x

⇒ -1 < \(\frac{sin x}x\leq1\)

⇒ -2 < \(\frac{2sin x}x\leq2\)

∴ \([\frac{2sinx}x]\) = \(\begin{cases}2&;x=0or sinx/x=1\\1&;1/2 \leq\frac{sin x}x<1\\0&;0\leq\frac{sin x}x<1/2\\-1&;\frac{sin x}x<0\end{cases}\)

\(\lim\limits_{x\to 0}f(x) = \lim\limits_{x\to 0}[\frac{2sin x}x]\) = 1

Discussion

6.	`lim_(xrarr0) ((1-cos2x)^2)/(2xtanx-xtan2x)` , isA. 2B. `-(1)/(2)`C. `(1)/(2)`D. `-2`
Answer» Correct Answer - D `lim_(xto0) ((1-cos2x)^2)/(2xtanx-xtan2x)` `=lim_(xto0) (4sin^4x)/(x(2tan x-(2tanx)/(1-tan^2x)))` ` =lim_(xto0) (4sin ^4x(1-tan^2x))/(-2tan^3x)` ` =-2 lim_(xto0) ((sinx)/(x))cos^3x(-tan^2x)=-2xx1xx1=-2`

Discussion

7.	The value of `lim_(xrarroo) ((x+3)/(x-1))^(x+1)` isA. eB. `e^2`C. `e^4`D. `1//e`
Answer» Correct Answer - C

Discussion

8.	`lim_(xrarroo) ((x+2)/(x+1))^(x+3)` is equal toA. 1B. eC. `e^2`D. `e^3`
Answer» Correct Answer - B

Discussion

9.	`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `(10)/(3)`B. `(3)/(10)`C. `(6)/(5)`D. `(5)/(6)`
Answer» Correct Answer - A We have, `lim_(xto0) ((1-cos2x)sin5x)/(x^2sin3x)` ` lim_(xto0) 2((sinx)/(x))^2xx((sin5x)/(5x))xx(5x)/(3x)xx((3x)/(sin3x))=(10)/(3)`

Discussion

10.	`lim_(x->0)((1-cos2x)sin5x)/(x^2sin3x)`A. `10//3`B. `3//10`C. `6//5`D. `5//6`
Answer» Correct Answer - A

Discussion

11.	If f(x) = { `sin[x] /[x],[x] != 0 ; 0, [x] = 0}` , Where[.] denotes the greatest integer function, then `lim_(x rarr 0) f(x)` is equal toA. 1B. 0C. -1D. none of these
Answer» Correct Answer - D We have, `[x]={(0", "0le x lt 1,),(0","-1lexlt0,):}` `f(x)={((sin(1))/(-1)sin1",when" -lexlt0,),(0" ,when "0lexlt1,):}` `rArr lim_(xto0^-) f(x)= sin 1 and, lim_(xto0^+)f(x)=0` `rArr lim_(xto0) f(x) ` does not exist.

Discussion

12.	`lim_(x->0)(sqrt((1-cos2x)/2))/x`A. 1B. -1C. 0D. none of these
Answer» Correct Answer - D `lim_(xto0)sqrt((1)/(2)(1-cos2x))/(x)=lim_(xto0) (\|sin\|)/(x)` Now, ` lim_(xto0^+)(\|sinx\|)/(x)=lim_(xto0) (sinx)/(x)=1` and , `lim_(xto0^-) (\|sinx\|)/(x)=lim_(xto0) -(sinx)/(x)=-1` Hence,`lim_(xto0) (\|sinx\|)/(x)` does not exist.

Discussion

13.	The value of `lim_(xto2)sqrt(1-cos 2(x-2))/(x-2)` , isA. `sqrt(2)`B. `-sqrt(2)`C. `2`D. none of these
Answer» Correct Answer - D We have `lim_(xto2)sqrt(1-cos 2(x-2))/(x-2)=lim_(xto2) (sqrt1-\|sin 2(x-2)\|)/(x-2)` Now, `lim_(xto2^-)sqrt(1-cos 2(x-2))/(x-2)=lim_(xto2^-) (sqrt1-\|sin 2(x-2)\|)/(x-2)` ` =-sqrt(2) lim_(xto2) (sin(x-2))/(x-2)[because x lt 2rArr x-2lt 0 rArr sin (x-2) lt 0rArr \|sin (x-2)\|=- sin (x-2)]` ` =-sqrt(2)` and, `lim_(xto2^+)sqrt(1-cos 2(x-2))/(x-2)=lim_(xto2^+) (\|sqrt2-sin (x-2)\|)/(x-2)` `=sqrt(2)lim_(xto2) (sin(x-2))/(x-2)=sqrt(2) [because x gt rArr x-2 gt0 rArr sin (x-2)gt 0 rArr \|sin (x-2)\|=sin (x-2)]` `therefore lim_(xto2^-) sqrt(1-cos 2(x-2))/(x-2)ne lim_(xto2^+) sqrt(1-cos 2(x-2))/(x-2)` `therefore lim_(xto2) sqrt(1-cos 2(x-2))/(x-2)` does not exist.

Discussion

14.	Evalaute`lim_(xto0) (x2^(x)-x)/(1-cosx)`
Answer» Correct Answer - `log4` `underset(xto0)lim(x(2^(x)-1))/(1-cosx)=underset(xto0)lim(2^(x)-1)/(x)xx(x^(2))/(1-cosx)` `=log2underset(xto0)lim(x^(2))/(2"sin"^(2)(x)/(2))` `=2log2=log4`

Discussion

15.	The value of `lim_(xto2) (sqrt(1+sqrt(2+x))-sqrt(3))/(x-2)" is "`
Answer» Correct Answer - A `underset(xto2)lim(sqrt(1+sqrt(2+x))-sqrt(3))/(x-2)` `underset(xto2)lim(1+sqrt(2+x)-3)/((sqrt(1+sqrt(2+x))+sqrt(3))(x-2))" "`(Rationalizing) `=underset(xto2)"lim"(sqrt(2+x)-2)/((sqrt(1+sqrt(2+x))+sqrt(3)")"(x-2)))` `=underset(xto2)"lim"((x-2))/("("sqrt(1+sqrt(2+x))+sqrt(3)")""("sqrt(2+x)+2")""("x-2")")` (Rationalizing) `=(1)/("("2sqrt(3")")4)=(1)/(8sqrt(3))`

Discussion

16.	Evaluate `lim_(xto2) sin(e^(x-2)-1)/(log(x-1))`
Answer» Correct Answer - 1 `underset(xto2)limsin(e^(x-2)-1)/(log(x-1))=underset(t to0)lim(sin(r^(t)-1))/(log(1+t))" "["Putting "x=2+t]` `=underset(t to0)lim(sin(e^(t)-1))/(e^(t)-1)xx(e^(t)-1)/(t)xx(t)/(log(1+t))` `=1xx1xx1=1`

Discussion

17.	Prove that `lim_(xto2) [x]` does not exists, where [.] represents the greatest integer function.
Answer» We know that greatest integer function changes its value at integral values of x. So, let us find LHL and RHL. `LHL=underset(xto2)lim[x]=underset(htoo)lim[2-h]` Sine h is positive and infinitely small `(2-h)` is value very close to 2 but smaller than 2 (say 1.99999) `:." "[2-h]=1,` when h approaches to zero. Thus, LHL=1 `RHL=underset(xto2)lim[x]=underset(hto0)lim[2+h]=[2.00001]=2` We observe that `LHLneRHL.` So, `underset(xto2)lim[x]` does not exist.

Discussion

18.	Evaluate `lim_(xto-2^(+)) (x^(2)-1)/(2x+4).`
Answer» Correct Answer - `oo.` `L= underset(xto-2^(+))lim(x^(2)-1)/(2x+4)` `=underset(hto0)lim((-2+h)^(2)-1)/(2(-2+h)+4)` `=underset(hto0)lim(3-4h+h^(2))/(2h)` For `hto0,` Numerator`to3` and Denominator `to0` `:." "Ltooo`

Discussion

19.	Evaluate `lim_(xto2) (x^(2)-x-2)/(x^(2)-2x-sin(x-2)).`
Answer» `underset(xto2)lim(x^(2)-x-2)/(x^(2)-2x-sin(x-2))" "`(0/0 from) `=underset(xto2)lim((x-2)(x+1))/((x-2)-sin(x-2))` `=underset(xto2)lim(x+1)/(x-(sin(x-2))/(x-2))` `(2+1)/(2-1)=3`

Discussion

20.	Evaluate `lim_(xto2) (sqrt((x+7))-3sqrt((2x-3)))/(root(3)((x+6))-2root(3)((3x-5))).`
Answer» We have L= `underset(xto2)lim(sqrt((x+7))-3sqrt((2x-3)))/(root(3)((x+6))-2root(3)((3x-5)))" "`(0/0 from) Let `x-2=t` such that when `xto2,t to0.` Then `L= underset(t to0)lim((t+9)^(1/2)-3(2t+1)^(1/3))/((t+8)^(1/3)-2(3t+1)^(1/3))" "`(0/0 from) `=3/2underset(t to0)lim((1+t/9)^(1/2)-(2t+1)^(1/2))/((1+t/8)^(1/3)-(3t+1)^(1/3))" "`(0/0 from) `=3/2underset(t to0)lim(1/2 t/9-(2t)1/2)/(t/8 1/3-(3t)1/3)=3/2(1/18-1)/(1/24-1)=34/23`

Discussion

21.	Evaluate `lim_(xto2^(+)) ([x-2])/(log(x-2)),` where `[.]` represents the greatest integer function.
Answer» `L=underset(xto2^(+))lim([x-2])/(log(x-2))` When `xto2^(+),x-2to0^(+)` or `[x-2]=0` Also, `log(x-2)tolog0^(+)to-oo` Thus, `L=("exact "0)/(-oo)=0`

Discussion

22.	Match the following lists (where `[x]` represents the greatest integer function) and then choose the correct code. Codes : `{:(,"a b c d"),((1),"s r q p"),((2),"q p s p"),((3), "s r p q"),((4),"p p q r"):}`
Answer» Correct Answer - `(2)` a. `underset(xto0)lims(-1)^([1//x])` `L.H.L.=underset(hto0)lim(0-h)(-1)^([(1)/(0-h)])=0` `R.H.L.=underset(hto0)lim(0+h)(-1)^([(1)/(0+h)])=0` b. `underset(xto2)lim(-1)^([x])` `L.H.L.=underset(hto0)lim(-1)^([2-h])=(-1)^(1)=-1` `R.H.L.=underset(hto0)lim(-1)^([2+h])=(-1)^(2)=1` So limit does not exist. c.`underset(xto(3)/(2))lim(x-[x])` `L.H.L.=underset(hto0)lim(3)/(2)-h-[(3)/(2)-h]=underset(hto0)lim(3)/(2)-h-1=(1)/(2)` `R.H.L.=underset(hto0)lim(3)/(2)+h-[(3)/(2)+h]=underset(hto0)lim(3)/(2)+h-1=(1)/(2)` `L.H.L.=R.H.L.=(1)/(2)` `underset(xto0)lim[x]((e^(1//x)-1)/(e^(1//x)+1))` `L.H.L.=underset(hto0)lim[0-h](((1)/(e^(0-h)-1))/((1)/(e^(0-h)+1)))` `=underset(hto0)lim[-h]((e^(-(1)/(h))-1)/((1)/(e^(0-h)+1)))=(-1)xx(-1)=1` R.H.L.`=underset(hto0)lim[0+h]((e^((1)/(h))-1)/(e^((1)/(h)+1)))=0` Limit does not exist

Discussion

23.	Given a real valued function f such that `f(x)={tan^2[x]/(x^2-[x]^2) , x lt 0 and 1 , x=0 and sqrt({x}cot{x}) , x lt 0` where [.] represents greatest integer function thenA. `p_(1)" ln "a_(1)+p_(2)" ln "a_(2)+...+p_(n)" ln "a_(n)`B. `a_(1)^(p_(1))+a_(2)^(p_(2))+...+a_(n)^(p_(n))`C. `a_(1)^(p_(1)).a_(2)^(p_(2))...a_(n)^(p_(n))`D. `sum_(r=1)^(n)a_(r)p_(r)`
Answer» Correct Answer - A::B::C::D We have `f(x)=underset(xto0^(+))lim(tan^(2){x})/((x^(2)-[x]^(2)))=underset(xto0^(+))lim(tan^(2)x)/(x^(2))=1" "(1)` `(becausexto0^(+),[x]=0implies{x}=x)` Also, `underset(xto0^(-))limf(x)=underset(xto0^(-))limsqrt({x}cot{x})=sqrt(cot1)" "(2)` `(becausexto0^(-),[x]=-1implies{x}=x+1implies{x}to1)` Also,`cot^(-1)(underset(xto0^(-))limf(x))^(2)=cot^(-1)(cot1)=1.` Also, `tan^(-1)(underset(xto0^(+))limf(x))=tan^(-1)1=(pi)/(4)`

Discussion

24.	If `f(x)=(3x^2+a x+a+1)/(x^2+x-2),`then which of the following can be correct`("lim")_(xvec1)f(x)e xi s t s a=-2``("lim")_(xvec-2)f(x)e xi s t s a=13``("lim")_(xvec1)f(x)=4/3``("lim")_(xvec-2)f(x)=-1/3`A. `"ln "a_(1)`B. `e^(a_(n))`C. a_(1)`D. `a_(n)`
Answer» Correct Answer - A::B::C::D `f(x)=(3x^(2)+ax+a+1)/((x+2)(x-1))` As `xto1,D^(r)to0," Hence as "xto 1,N^(r)to0.` Therefore, `3+2a+1=0" or "a=-2` As `xto-2,D^(r)to0." Hence as "xto-2,N^(r)to0.` Therefore, `12-2a+a+1=0" or "a=13` Now, `underset(xto1)limf(x)=underset(xto1)lim(3x^(2)-2x-1)/((x+2)(x-1))=underset(xto1)lim((3x+1)(x-1))/((x+2)(x-1))=(4)/(3)` Now, `underset(xto-2)lim(3x^(2)+13x+14)/((x+2)(x-1))=underset(xto-2)lim((3x+7)(x+2))/((x+2)(x-1))=-(1)/(3)`

Discussion

25.	`A_(i)=(x-a_(i))/(\|x-a_(i)\|),i=1,2,...,n," and "a_(1)lta_(2)lta_(3)lt...lta_(n).` If `1lemlen,minN,` then `lim_(xtoa_(m)) (A_(1)A_(2)...A_(n))`
Answer» Correct Answer - C We have `A_(i)=(x-a_(i))/(\|x-a_(i)\|),i=1,2,...n,` and `a_(1)lta_(2)lt...lta_(n-1)lta_(n).` Let x be in the left neighborhood of `a_(m).` Then `x-a_(i)lt0" for "i=m,m+1,…,n` and `x-a_(i)lt0" for "i=1,2,…,m-1` and `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1 " for "i=m,m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1 " for "i=1,2,...,m-1` Similarly, if x is in the right neighborhood of `a_(m)`, then `x=a_(i)lt0" for "i=m+1,...,n," and "x-a_(i)lt0" for "i=1,2,...,m.` Therefore, `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1" for "i=m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1" for "i=1,2,...,m` Now, Now, `underset(xtoa_(m)^(-))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m+1)` and `underset(xtoa_(m)^(+))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m)` Hence, `underset(xtoa_(m))lim(A_(1)A_(2)...A_(n))` does not exist.

Discussion

26.	`A_(i)=(x-a_(i))/(\|x-a_(i)\|),i=1,2,...,n," and "a_(1)lta_(2)lta_(3)lt...lta_(n).` If `1lemlen,minN,` then the value of `R=lim_(xtoa_(m)+) (A_(1)A_(2)...A_(n))`isA. `e^(-(1)/(4))`B. `e^(-(1)/(2))`C. `e^(-2)`D. `e^(-4)`
Answer» Correct Answer - D We have `A_(i)=(x-a_(i))/(\|x-a_(i)\|),i=1,2,...n,` and `a_(1)lta_(2)lt...lta_(n-1)lta_(n).` Let x be in the left neighborhood of `a_(m).` Then `x-a_(i)lt0" for "i=m,m+1,…,n` and `x-a_(i)lt0" for "i=1,2,…,m-1` and `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1 " for "i=m,m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1 " for "i=1,2,...,m-1` Similarly, if x is in the right neighborhood of `a_(m)`, then `x=a_(i)lt0" for "i=m+1,...,n," and "x-a_(i)lt0" for "i=1,2,...,m.` Therefore, `A_(i)=(x-a_(i))/(-(x-a_(i)))=-1" for "i=m+1,...,n` and `A_(i)=(x-a_(i))/(x-a_(i))=1" for "i=1,2,...,m` Now, Now, `underset(xtoa_(m)^(-))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m+1)` and `underset(xtoa_(m)^(+))lim(A_(1)A_(2)...A_(n))=(-1)^(n-m)` Hence, `underset(xtoa_(m))lim(A_(1)A_(2)...A_(n))` does not exist.

Discussion

27.	Evaluate the following limits:\(\lim\limits_{x\longrightarrow2} [\frac {x^{-3}-2^{-3}}{x-2}]\)lim [x-3-2-3/ x-2] (x ∈2)
Answer» \(\lim\limits_{x\longrightarrow2} \frac {x^{-3}-2^{-3}}{x-2} \)= (-3). (2)^-4 ^...[∴\(\lim\limits_{x\longrightarrow\,a} \frac {x^{n}-2^{n}}{x-a} = n.a^{n-1}\) = \(-3 \times \frac {1}{2^4} = \frac {-3}{16}\)

27.

Evaluate the following limits:\(\lim\limits_{x\longrightarrow2} [\frac {x^{-3}-2^{-3}}{x-2}]\)lim [x-3-2-3/ x-2] (x ∈2)

Answer»

\(\lim\limits_{x\longrightarrow2} \frac {x^{-3}-2^{-3}}{x-2} \)= (-3). (2)^-4

^...[∴\(\lim\limits_{x\longrightarrow\,a} \frac {x^{n}-2^{n}}{x-a} = n.a^{n-1}\)

= \(-3 \times \frac {1}{2^4} = \frac {-3}{16}\)

Discussion

28.	Find the derivative of 99x at x = 100.
Answer» We have, f'(100) = lim_(h →0) (f(100 + h) - f(100))/h = lim_(h →0) (99(100 + h) - 99(100))/h = lim_(h→0) 99h/h = 99

28.

Find the derivative of 99x at x = 100.

Answer»

We have,

f'(100) = lim_(h →0) (f(100 + h) - f(100))/h

= lim_(h →0) (99(100 + h) - 99(100))/h

= lim_(h→0) 99h/h = 99

Discussion

29.	Evaluate `lim_(x rarr 2) (2x-2)`.
Answer» Correct Answer - 2

Discussion

30.	_______ helps in finding the areas of the curves.
Answer» Correct Answer - integration

Discussion

31.	The radius of circle tends to zero then it approaches a ________.
Answer» Correct Answer - point

Discussion

32.	A circle is inscribed in a polygon. As the number of sides increases, the difference in areas of circle and polygon _________.
Answer» Correct Answer - decreases

Discussion

33.	When the number of sides of a polygon tends to infinity, it approaches _______.
Answer» Correct Answer - circle

Discussion

34.	Find lim(x→2) (1/x + 1/2)/(x + 2)
Answer» lim_(x→2)(1/x + 1/2)/(x + 2) = (1/2 + 1/2)/(2 + 2) = 1/4

Discussion

35.	Find lim(x→0) (ax + b)/(cx + 1)
Answer» lim_(x→0) (ax + b)/(cx + 1) = (a(0) + b)/((0) + 1) = b

Discussion

36.	Find lim(x→2) f(x), where f(x) = 3x
Answer» LHL = lim_(x→2^-) 3x = 6 RHL = lim_(x→2⁺) 3x = 6 ∴ LHL = RHL = 6 ∴ lim_(x_→₂₎ f(x) = 6

36.

Find lim(x→2) f(x), where f(x) = 3x

Answer»

LHL = lim_(x→2^-) 3x = 6

RHL = lim_(x→2⁺) 3x = 6

∴ LHL = RHL = 6

∴ lim_(x_→₂₎ f(x) = 6

Discussion

37.	Find lim(x→2) f(x), where f(x) = 3 a constant function
Answer» LHL = lim_(x→2^-) 3 = 3 RHL = lim_(x→2⁺) 3 = 3 ∴ LHL = RHL = 3 ∴ lim_(x→3) f(x) = 3

37.

Find lim(x→2) f(x), where f(x) = 3 a constant function

Answer»

LHL = lim_(x→2^-) 3 = 3

RHL = lim_(x→2⁺) 3 = 3

∴ LHL = RHL = 3

∴ lim_(x→3) f(x) = 3

Discussion

38.	Evaluate: lim(x→π) (sin(π - x))/(π(π - x))
Answer» lim_(x→π) (sin(π - x))/(π(π - x)) = lim(y→0) siny/πy, where y = π- x = 1/π

38.

Evaluate: lim(x→π) (sin(π - x))/(π(π - x))

Answer»

lim_(x→π) (sin(π - x))/(π(π - x)) = lim(y→0) siny/πy, where

y = π- x

= 1/π

Discussion

39.	Evaluate: lim(x→0) cosx/(π - x)
Answer» lim_(x→0) cosx/(π - x) = cos0/(π - 0) = 1/π

Discussion

40.	Solve: \(\lim\limits_{x \to0}\frac{sin4x}{sin2x}\)
Answer» = \(\frac{\lim\limits_{x \to 0}\frac{sin4x}{4x} . 4x}{\lim\limits_{x\to0}\frac{sin2x}{2x}.2x}\) = \(\frac{1\times 4}{1 \times2} = 2\)

40.

Solve: \(\lim\limits_{x \to0}\frac{sin4x}{sin2x}\)

Answer»

= \(\frac{\lim\limits_{x \to 0}\frac{sin4x}{4x} . 4x}{\lim\limits_{x\to0}\frac{sin2x}{2x}.2x}\)

= \(\frac{1\times 4}{1 \times2} = 2\)

Discussion

41.	Evaluate the following limit : \(\lim\limits_{\text x \to0}\cfrac{tan\,2\text x-sin\,2\text x}{\text x^3}\)lim(x→0) (tan 2x - sin 2x)/x3
Answer» \(\lim\limits_{\text x \to0}\cfrac{tan\,2\text x-sin\,2\text x}{\text x^3} \) lim(x→0) (tan 2x - sin 2x)/x³ To Find: Limits NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form. In this Case, indeterminate Form is \(\cfrac00\) \(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}\) \(=\underset{x\to0}{lim}\frac{sin\,2x-sin\,2x\,cos\,2x}{x^3}\) (Multiplying numerator by cos 2x because \(\underset{x\to0}{lim}\) cos 2x = 1) \(=\underset{x\to0}{lim}\frac{sin\,2x(1-cos\,2x)}{x^3}\) \(=\underset{x\to0}{lim}\frac{2sin\,2x}{2x}\times\frac{4(1-cos\,2x)}{(2x)^2}=4\) Therefore, \(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}=4\) Other method: - \(\underset{x\to0}{lim}\) \(\frac{tan\,2x-sin\,2x}{x^3}\) \((\frac00)\) \(=\underset{x\to0}{lim}\) \(\frac{2sec^2\,2x-2cos\,2x}{3x^2}\) \((\frac00)\) \(=\underset{x\to0}{lim}\) \(\frac{8sec^22x\,tan\,2x+4\,sin\,2x}{6x}\) \(=\frac83\,\underset{x\to0}{lim}\,sec^22x\,\underset{x\to0}{lim}\frac{tan\,2x}{2x}+\frac{4}{3}\,\underset{x\to0}{lim}\frac{sin\,2x}{2x}\) \(=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4\)

41.

Evaluate the following limit : \(\lim\limits_{\text x \to0}\cfrac{tan\,2\text x-sin\,2\text x}{\text x^3}\)lim(x→0) (tan 2x - sin 2x)/x3

Answer»

\(\lim\limits_{\text x \to0}\cfrac{tan\,2\text x-sin\,2\text x}{\text x^3} \)

lim(x→0) (tan 2x - sin 2x)/x³

To Find: Limits

NOTE: First Check the form of imit. Used this method if the limit is satisfying any one from 7 indeterminate form.

In this Case, indeterminate Form is \(\cfrac00\)

\(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}\) \(=\underset{x\to0}{lim}\frac{sin\,2x-sin\,2x\,cos\,2x}{x^3}\) (Multiplying numerator by cos 2x because \(\underset{x\to0}{lim}\) cos 2x = 1)

\(=\underset{x\to0}{lim}\frac{sin\,2x(1-cos\,2x)}{x^3}\) \(=\underset{x\to0}{lim}\frac{2sin\,2x}{2x}\times\frac{4(1-cos\,2x)}{(2x)^2}=4\)

Therefore, \(\underset{x\to0}{lim}\frac{tan\,2x-sin\,2x}{x^3}=4\)

Other method: - \(\underset{x\to0}{lim}\) \(\frac{tan\,2x-sin\,2x}{x^3}\) \((\frac00)\)

\(=\underset{x\to0}{lim}\) \(\frac{2sec^2\,2x-2cos\,2x}{3x^2}\) \((\frac00)\)

\(=\underset{x\to0}{lim}\) \(\frac{8sec^22x\,tan\,2x+4\,sin\,2x}{6x}\)

\(=\frac83\,\underset{x\to0}{lim}\,sec^22x\,\underset{x\to0}{lim}\frac{tan\,2x}{2x}+\frac{4}{3}\,\underset{x\to0}{lim}\frac{sin\,2x}{2x}\)

\(=\frac{8}{3}+\frac{4}{3}=\frac{12}{3}=4\)

Discussion

42.	Evaluate `lim_(ntooo) (-1)^(n-1)sin(pisqrt(n^(2)+0.5n+1)),"where "nin N`
Answer» L=`(-1)^(n-1)underset(ntooo)limsin(pisqrt(n^(2)+0.5n+1))` `=underset(ntooo)lim(-1)^(n-1)(-1)^(n-1)sin(npi-pisqrt(n^(2)+n/2+1))` `=underset(ntooo)limsinpi{{(n-sqrt(n^(2)+n/2+1))(n+sqrt(n^(2)+n/2+1)))/(n+sqrt(n^(2)+n/2+1))}` `=underset(ntooo)limsinpi{(n^(2)-n^(2)-n/2-1)/(n(1+sqrt(1+1/2n+1/n^(2)))}}` `=underset(ntooo)limsinpi{(-n/2-1)/(n(1+sqrt(1+(1)/(2n)-(1)/(n^(2))))}}` `=underset(ntooo)limsinpi((-1/2-1/n)/(1+sqrt(1+(1)/(2n)+(1)/(n^(2)))))=sin(-(pi)/(4))=-(1)/(sqrt(2))`

Discussion

43.	Evaluate `lim_(ntooo) (n^(p)sin^(2)(n!))/(n+1),"wher "0ltplt1.`
Answer» We have `underset(ntooo)lim(n^(p)sin^(2)(n!))/(n+1)" "((oo)/(oo)"from")` `=underset(ntooo)lim(sin^(2)(n!))/(n^(1-p)(1+1/n))` `=("some number between 0 and1")/(ooxx1)=0" "("as "0ltplt1)`

Discussion

44.	If `{(sin{cosx})/(x-pi/2),x!=pi/2 and 1,x=pi/2,` where {.} represents the fractional part function, then `f(x)` isA. `-1`B. 1C. non-existantD. none of these
Answer» Correct Answer - C We have ` lim_(xtopi//2^-)f(x)=lim_(xto pi//2^-)(sin{cosx})/(x-pi//2)` ` rArr lim_(xtopi//2^-)f(x)=lim_(hto0)(sin{cos((pi)/(2)-h)})/(pi//2-h-pi//2)=lim_(hto0) (sin{sin h})/(-h)` ` rArr lim_(xtopi//2^-)f(x)=lim_(hto0) (sin(sin h))/(-h)=-lim_(hto0) (sin(sin h))/(sin h)xx(sinh)/(h)=-1` and, ` lim_(xtopi//2^+)f(x)=lim_(hto0) (sin{-sin h})/(x-pi//2)=lim_(hto0) (sin{cos (pi//2+h)})/((pi)/(2)+h-(pi)/(2))` ` rArr lim_(xtopi//2^+)f(x)=lim_(hto0) (sin{-sinh})/(h)=lim_(hto0) (sin(1-sinh))/(h)to oo` Hence, `lim_(xto pi//2)f(x)` does not exist.

Discussion

45.	The value of `lim_(xrarr oo) 1+(1)/(x^n)^x,ngt 0`, isA. `1, if n lt 1`B. `1,if n gt 1`C. `e,if n gt 1`D. `e,if n lt 1`
Answer» Correct Answer - B We have , `lim_(xto oo) (1+(1)/x^n)^x=e^(lim_(xtooo)(x)/(x^n))=e^(lim_(xto oo)x^(1-n)= {(e^(0)=1,,"if " ngt1,,),(e,,"if " n=1,,),(oo,,"if " nlt1,,):}`

Discussion

46.	`lim_(xrarr oo) (n^p sin^2(n!))/(n+1),0ltplt1`, is equal to
Answer» Correct Answer - A We have , `lim_(xto oo) (n^p sin^2(n!))/(n+1)=lim_(ntooo)(sin^2(n!))/(1+(1)/(n))xxn^(p-1)` `lim_(xto oo) (n^p sin^2(n!))/(n+1)=lim_(ntooo)(sin^2(n!))/(n^(1-p))xx(1)/(1+(1)/(n))` ` rArr lim_(xto oo) (n^p sin^2(n!))/(n+1)=("An oscillating number")/(oo) xx1=0xx1=0`.

Discussion

47.	`lim(x->a_-) {(\|x\|^3)/a-[x/a]^3} ,(a < 0)`, where `[x]` denotes the greatest integer less than or equal to `x` is equal to:A. `a^2-3`B. `a^2-1`C. `a^2`D. none of these
Answer» Correct Answer - C When `xtoa^-1`, we have `0lt x lt a rArr 0lt (x)/(a)lt 1rArr [(x)/(a)] =0` ` therefore lim-(xtoa^-){\|x\|^3/(a)-[(x)/(a)]^3}=lim_(xtoa^-)(\|x\|^3)/(a)=(a^3)/(a)=a^2`

Discussion

48.	In the following examples, given ∈ > 0, find a δ > 0 such that whenever, \|x – a\| < δ, we must have \|f(x) – l\| < ∈.\(\lim\limits_{x\longrightarrow2} (2x+3)=7 \)lim (2x +3) = 7 (x ∈ 2)
Answer» We have to find some δ so that \(\lim\limits_{x\longrightarrow2} (2x+3)=7 \) Here a = 2, l = 1 and f(x) = 2x + 3 Consider ∈ > 0 and \|f(x) – l\| < ∈ ∴ \|(2x + 3) – 7\| < ∈ ∴ \|2x + 4\| < ∈ ∴ 2(x – 2)\|< ∈ ∴ \|x – 2\| < ∈/2 ∴ δ ≤ ∈/2 such that \|2x + 4\| < δ ⇒ \|f(x) – 7\| < ∈

48.

In the following examples, given ∈ > 0, find a δ > 0 such that whenever, |x – a| < δ, we must have |f(x) – l| < ∈.\(\lim\limits_{x\longrightarrow2} (2x+3)=7 \)lim (2x +3) = 7 (x ∈ 2)

Answer»

We have to find some δ so that

\(\lim\limits_{x\longrightarrow2} (2x+3)=7 \)

Here a = 2, l = 1 and f(x) = 2x + 3

Consider ∈ > 0 and |f(x) – l| < ∈

∴ |(2x + 3) – 7| < ∈

∴ |2x + 4| < ∈

∴ 2(x – 2)|< ∈

∴ |x – 2| < ∈/2

∴ δ ≤ ∈/2 such that

|2x + 4| < δ ⇒ |f(x) – 7| < ∈

Discussion

49.	Evaluate `lim_(xtooo) 2^(x-1)tan((a)/(2^(x))).`
Answer» Since, `underset(xtooo)lim(a)/(2^(x))=0,` we have `underset(xtooo)lim2^(x-1)tan((a)/(2^(x)))" "`(`ooxx0` form) `=underset(xtooo)lim(a)/(2).tan((a)/(2^(x)))/(((a)/(2^(x))))" "`(0/0 from)`=a/2xx1=a/2`

Discussion

50.	Evaluate `lim_(x to 0) x[tan^(-1)((x+1)/(x+2))-tan^(-1)((x)/(x+2))].`
Answer» `underset(xtooo)limx["tan"^(-1)(x+1)/(x+2)-"tan"^(-1)(x)/(x+2)]` `=underset(xtooo)limxtan^(-1)(((x+1)/(x+2)-(x)/(x+2))/(1+(x+1)/(x+2).(x)/(x+2)))` `=underset(xtooo)limxtan^(-1)((x+2)/(2x^(2)+5x+4))` `=underset(xtooo)lim(tan^(-)((x+2)/(2x^(2)+5x+4))/((x+2)/(2x^(2)+5x+4)))xx(x(x+2))/(2x^(2)+5x+4)` `=1xx1/2=1/2`

Discussion

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