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101.

Define limit of a function.

Answer»

Limit of a function: When the value of a variable x Is brought closer and closer to a number ‘a’, the value of function f(x) approaches closer and closer to a definite number ‘l’, then we can say that as x tends to ‘a’ f(x) tends to ‘l’ that is x → a, f(x) → 1. This definite number ‘l’ is called limit of a function f(x). Symbolically it can be written as follows:

\(\lim\limits_{x\to a}\,f(x)=l\)

‘l’ is the limit of f(x). Hence f(x) ≠ l.
Discussion
102.

State division working rule of limit.

Answer»

f (x) and g (x) are two functions of real variable x and \(\lim\limits_{x\to a}\) f(x) = 1, \(\lim\limits_{x\to a}\) g (x) = m.

If \(\frac{f(x)}{g(x)}\) is the division of the two functions. then division working rule of limit is as follows:

\(\lim\limits_{x\to a}\,\left[\frac{f(x)}{g}(x)\right]\\=\frac{\lim\limits_{x\to a}\,f(x)}{\lim\limits_{x\to a}\,g(x)}\\=\frac{l}{m},m\neq0\)

Thus, the limit of division of two functions is equal to the division of their limits, where the limit of the function in denominator is not zero.
Discussion
103.

find limit by Tabular Method .

Answer»

To find limit by Tabular Method :

Limiting Value of xIncreasing ValuesDecreasing Values
-2-2.1. -2.01, -2.001, -2.0001,…– 1.9, – 1.99, -1.999, – 1.9999,…
– 1-1.1, -1.01, -1.001, -1.0001,…-0.9, -0.99, -0.999, -0.9999,…
0-0.1, -0.01, -0.001, -0.0001….0.1, 0.01, 0.001, 0.0001,…
10.9, 0.99, 0.999, 0.9999,…1.1, 1.01, 1.001, 1.0001,…
21.9, 1.99, 1.999, 1.9999,…2.1, 2.01, 2.001, 2.0001,…
Discussion
104.

 Punctured 8 neighbourhood of a.

Answer»
  • N * (a, δ) = {x | (a – δ) < x < (a + δ), x ≠ a, x∈ R}
  • N (a, δ) = |x – a|< δ (Modulus form)
  • N (a, δ) = (a – δ, a + δ) (Interval form)
Discussion
105.

δ neighbourhood of a .

Answer»

N (a, δ) = (x| (a – δ) < x < (a + δ), x ∈ R}
Discussion
106.

For each positive integer n, let ` y_(n) = 1/n ((n+1)(n+2)…(n+n))^(1/n)`. For ` x in R`, let [x] be the greatest integer less than or equal to x. If ` lim_( n to infty) y_(n) = L`, then the value of [L] is …………….. .

Answer» Correct Answer - `(1)`
We have,
` y_(n)=1/n [( n+ 1)(n+1)(n+2)...(n+n)]^(1//n) and underset( n to infty) lim y_(n) = L `
` rArr L = underset( n to infty) lim 1/n [(n+1)(n+2)(n+3)...(n+n)]^(1//n) `
` rArr L = underset( n to infty) lim [( 1+1/n)(1+2/n)(1+3/n)...(1+n/n)]^(1/n)`
`rArr log L = underset( n to infty) lim 1/n [log (1+1/n)+ log (1+2/n) ...log(1+n/n)]`
` rArr log L = underset( n to infty) lim 1/n underset( r =1) overset( n) sum log (1+ r/n)`
`rArr log L = int_(0)^(1) underset"II" 1 xx underset("I")log(1+x) dx `
` rArr log L = (x* log (1+ x))_(0)^(1) -int_(0)^(1) [d/(dx)(log(1+x))int dx]dx`
[by using integration by parts]
`rArr log L = [ x log (1+x)]_(0)^(1) - int _(0)^(1) x/(1+x) dx `
` rArr log L = log 2 - int_(0)^(1) ((x+1)/(x+1) -1/(x+1)) dx`
` rArr log L = log 2 - [x]_(0)^(1) + [log(x+1)]_(0)^(1)`
` rArr log L = log 2 -1 + log 2 -0`
` rArr log L = log 4 - log e = log e = log 4/e rArr L = 4/e rArr [L] = [4/e] = 1`
Discussion
107.

If N (k1, 0.5) = (19.5, k2), then find the values of k1 and k2.

Answer»

N (k1, 0.5) = (19.5, k2)

k1 – <5 = 19.5, k1 + S = k2

Here, δ = 0.5

∴ k1 – 0.5 = 19.5

∴ k1 = 19.5 + 0.5 = 20

Putting k1 – 20 and δ = 0.5 in k1 + δ = k2,

20 + 0.5 = k2

k2 = 20.5

Hence, k1 = 20, k2 = 20.5
Discussion
108.

Let `[.]` represent the greatest integer function and `f (x)=[tan^2 x]` then :

Answer» Correct Answer - B
Given, f(x) = `[tan^(2) x]`
Now, ` - 45^(@) lt x lt 45^(@) `
`rArr" " tan (-45^(@)) lt tan x lt tan (45^(@))`
` rArr" "-tan 45^(@) lt tan x lt tan (45^(@))`
`rArr" " 0 lt tan^(2) x lt 1`
`rArr " "[tan^(2) x] = 0`
i.e. f(x) is zero for all values of x from ` x =- 45^(@) to 45^(@)`. Thus, f(x) exists when ` x to 0 ` and also it is continuous at x = 0. Also, f(x) is differentiable at x = 0 and has a value of zero.
Therefore, (b) is the answer.
Discussion
109.

If ` f (x) = x (sqrtx+sqrt((x+1))`, thenA. f (x) is continuous but not differentiable at x = 0B. f(x) is differentiable at x = 0C. f(x) is not differentiable at x = 0D. None of the above

Answer» Correct Answer - C
Given,` f(x) = x (sqrtx+sqrt(x+1)`
`rArr f(x) ` would exists when ` x ge 0`.
`:.` f(x) is not continous at x = 0 ,
because LHL does not exist. ltbr. Hence, option (c ) is correct.
Discussion
110.

Let [x] denotes the greatest integer less than or equal to x. If `f(x) = [x sin pix]`, then f(x) isA. continuous at x = 0B. continuous in `(-1, 0)`C. differentiable at x = 1D. differentiable in `(-1, 1)`

Answer» Correct Answer - A::B::D
We have, for ` -1 lt x lt 1`
` rArr" " [x tan pix] = 0`
Also, x sin ` pi x` becomes negative and numerically less than 1 when x is slightly greater than 1 and so by definition of [x] .
` f(x) = [ x sin pi x ] =- 1, " when " 1 lt x lt 1 + h`
Thus, f(x) is constant and equal to 0 in the closed interval `[-1, 1] ` and so f(x) is continuous and differentiable in the open interval ` (-1, 1)`.
At x = 1, f(x) is discontinuous, since ` underset( h to 0) lim (1-h) = 0 `
and `underset( h to 0) lim (1+h) =- 1`
`:.` f(x) is not differentiable at x = 1 .
Hence, (a), (b) and (d) are correct answers.
Discussion
111.

The function `f(x) = [x] cos((2x-1)/2) pi` where [ ] denotes the greatest integer function, is discontinuousA. all xB. all integer pointsC. no xD. x which is not an integer

Answer» Correct Answer - C
Here, `f(x) = [x] cos ((2x -1)/2) pi`
f(x)=`:." "f(x)={{:(-cos((2x-1)/2)pi" ,"-1 le x lt 0),(0" , "0 le x lt 1),(cos((2x-1)/2) pi" , " 1 le x lt 2),(2 cos ((2x -1)/2) pi" , "2le x lt 3):}`
which shows RHL = LHL at x = n `in ` Integer as if x = 1
`rArr underset( x to 1^(+)) lim cos((2x-1)/2) pi = 0 and underset ( x to 1^(-)) lim 0= 0 `
Also, f(1) = 0
`:.` Continous at x = 1.
Similarly, when x = 2,
` underset( x to 2^(+)) lim f(x) = underset( x to 2^(-)) lim f(x) = 0 `
Thus, function is discontinuous at no x.
Hence, option (c ) is the correct answer.
Discussion
112.

If `f(x) ={{:(sin[x]/[[x]]", "[x] ne 0),(" "0", "[x] ne 0):}` where, [x] denotes the greatest integer less than or equal to x, then ` lim_(x to 0) f(x) ` equalsA. 1B. 0C. `-1`D. None of these

Answer» Correct Answer - D
Since, `f(x) = {{:((sin [x])/[[x]]", "[x] ne 0 ),("0, "[x] = 0):}`
`rArr f(x) = {{:((sin [x])/[[x]]", "x inR-[0,1)),("0, "0 le x lt 1):}`
At x = 0,
RHL = ` underset(x to 0^(+)) lim 0 = 0`
and LHL = `underset(x to 0^(-)) lim (sin [x])/[x] = underset( h to 0) lim (sin [0-h])/([0-h]) `
` = underset( h to 0) lim (sin (-1))/(-1) = sin 1 `
`:. ` Limit does not exist.
Discussion
113.

In the following, [x] denotes the greatest integer less than or equal to x. `{:(,"Column I",,"Column II"),(A.,x|x|,p.,"continuous in (-1, 1)"),(B.,sqrt|x|,q.,"differentiable in (-1, 1) "),(C.,x+[x],r.,"strictly increasing (-1, 1)"),(D.,|x-1|+|x+1| " in (-1,1)",s.,"not differentiable atleast at one point in (-1, 1)"):}`

Answer» Correct Answer - `(A) to p,q,r,s; (B) to p, s; (D) to p, s `
A. x|x| is continuous, differentiable and strictly increasing in (-1, 1).
B. `sqrt|x|` is continuous in (-1, 1) and not differentiable at x = 0.
C. x+[x] is strictly increasing in (-1, 1) and discontinuous at x = 0
`rArr` not differentiable at x = 0.
D. `|x-1|+|x+1|=2" in "(-1, 1)`
`rArr ` The function is continuous and differentiable in (-1, 1) .
Discussion
114.

The value of `lim_(x rarr 0)(a^(2x)-1)/(2x)` isA. `(1)/(4)log_(e)a`B. 1C. `(1)/(2)`D. `log_(e)a`

Answer» Correct Answer - D
Discussion
115.

The value of `lim_(x rarr 0) (e^(x2)-1)/(x)` isA. 1B. 0C. does not existD. nono of these

Answer» Correct Answer - B
Discussion
116.

The value of `lim_(xrarroo)(1+(1)/(x))^(x)` isA. 1B. eC. `e^(-1)`D. does not exist

Answer» Correct Answer - B
Discussion
117.

Evaluate the following limits: `lim_(xrarr0)(x cot2x)`

Answer» Correct Answer - `1/2`
Given limit `=1/2xxlim_(2xto0)(2x)/(tan2x)=((1)/(2)xx1)=1/2.`
Discussion
118.

Evaluate the following limits: `lim_(xrarr0)((cot2x-cosecex))/(x)`

Answer» Correct Answer - `-1`
Given limit `=lim_(xto0)((cos2x-1))/(xsin2x)=lim_(xto0)(-(1-cos2x))/(xsin2x)`
`=lim_(xto0)(-2sin^(2)x)/(2xsinxcosx)=-lim_(xto0)(tanx)/(x)=-1.`
Discussion
119.

Let `f(x){{:((x)/(|x|)",",xne0),(0",",x=0):}` Show that `lim_(xrarr0)f(x)` does not exist.

Answer» `lim_(xto0^(+))f(x)=lim_(hto0)(0+h)=lim_(hto0)f(h)=lim_(hto0|-h|)=lim_(hto0)h/h=1" "[becausehgt0]`
`lim_(xto0^(-))f(x)=lim_(hto0)f(0-h)=lim_(hto0)(-h)/(|-h|)=lim_(hto0)(-h)/(h)=-1.`
`thereforelim_(xto0^(+))f(x)nelim_(xto0)f(x)and so limf(x)` does not exist.
Discussion
120.

Evaluate `lim_(xrarra)((x+2)^(3//2)-(a+2)^(3//2))/(x-a).`

Answer» `lim_(xtoa)((x+2)^(3//2)-(a+2)^(3//2))/(x-a)`
`=underset((x+2)to(a+2))(lim)((x+2)^(3//2)-(a+2)^(3//2))/((x+2)-(a+2))`
`=3/2.(a+2)^(((3)/(2)-1))=3/2(a+2)^(1//2)" "[becauselim_(xtoa)((x^(n)-a^(n))/(x-a))=na^(n-1)].`
Discussion
121.

`lim_(x->0)(sin3x+7x)/(4x+sin2x)`

Answer» `lim_(xto0)((sin3x+7x)/(4x+sin2x))=lim_(xto0)(((sin3x)/(x)+7)/(4+(sin2x)/(x))) " "["dividing num. and denom. by x"]`
`=lim_(xto0){((sin3x)/(3x).3+7)/(4+((sin2x)/(2x)).2)}=([3xxlim_(3xto0)(sin3x)/(3x)]+7)/(4+[2xxlim_(2xto0)((sin2x)/(2x))])`
`=((3xx1)+7)/(4+(2xx1))=10/6=5/3.`
Discussion
122.

Evaluate `lim_(xrarra){(x^(12)-a^(12))/(x-a)}.`

Answer» `lim_(xtoa){(x^(12)-a^(12))/(x-a)}=12a^((12-1))=12a^(11)" "[becauselim_(xtoa)((x^(n)-a^(n))/(a-x))=na^(n-1)].`
Discussion
123.

Evalute `lim_(xrarr1) (x^(15)-1)/(x^(10)-1)`

Answer» `lim_(xto1)(x^(15)-1)/(x^(10)-1)=lim_(xto1){((x^(15)-1)/(x-1))div((x^(10)-1)/(x-1))}`
`=lim_(xto1)((x^(15)-1)/(x-1))divlim_(xto1)((x^(10)-1)/(x-1))`
`implies(15xx10^(14))div(10xx1^(9))=15/10=3/2.`
Discussion
124.

If `lim_(x to -a)(x^(7)+a^7)/(x+a)=7` then the value of a isA. 1B. -1C. 7D. -7

Answer» Correct Answer - A,B,D
Discussion
125.

Let `f(x) (1-cos4x)/(x^(2)),g(x)=(sqrtx)/((sqrt(16+sqrtx))-4)and q(x)=(e^(2x)-x^(x)+1)/(x^(2x+e^(x)+1)),(x in R).` `lim_(x to 0)g(x)=`A. `(1)/(8)`B. 8C. 2D. `(1)/(2)`

Answer» Correct Answer - B
Discussion
126.

Let `f(x) (1-cos4x)/(x^(2)),g(x)=(sqrtx)/((sqrt(16+sqrtx))-4)and q(x)=(e^(2x)-x^(x)+1)/(x^(2x+e^(x)+1)),(x in R).` `lim_(xrarr0)f(x)`A. `(1)/(2)`B. 2C. `(1)/(8)`D. 8

Answer» Correct Answer - D
Discussion
127.

Evaluate `lim_(xrarr0)(xtan4x)/(1-cos4x).`

Answer» `lim_(xto0)(xtan4x)/(1-cosx)=lim_(xto0)(xsin4x)/(cos4x(2sin^(2)2x))`
`=lim_(xto0)(2xsin2xcos2x)/(cos4x(2sin^(2)2x))=lim_(xto0)((cos2x)/(cos4x).(2x)/(sin2x)xx(1)/(2))`
`=1/2xx(lim_(2xto0)cos2x)/(lim_(4xto0)cos4x)xxlim_(2xto0)((2x)/(sin2x))=((1)/(2)xx(1)/(1)xx1)=1/2.`
Discussion
128.

Evaluate `lim_(xrarr(pi)/(4))((sinx-cosx))/((x-(pi)/(4))).`

Answer» Put `(x-(pi)/(4))=y"so that when"y to (pi)/(4) then y to 0.`
`thereforelim_(xto(pi)/(4))((sinx-cosx))/((x-(pi)/(4)))`
`=lim_(yto0)({sin""((pi)/(4)+y)-cos((pi)/(4)+y)})/(y)" "["putting""("x-(pi)/(4)")"=y]`
`=lim_(yto0)({(sin""(pi)/(4)cosy+cos""(pi)/(4)siny)-(cos""(pi)/(4)cosy-sin""(pi)/(4)siny)})/(y)`
`=(2)/(sqrt2)xxlim_(yto0)((siny)/(y))=(2sqrtxx1)=sqrt2.`
Discussion
129.

Let `f(x) (1-cos4x)/(x^(2)),g(x)=(sqrtx)/((sqrt(16+sqrtx))-4)and q(x)=(e^(2x)-x^(x)+1)/(x^(2x+e^(x)+1)),(x in R).` `lim_(x to 0)q(x)=`

Answer» Correct Answer - B
Discussion
130.

Evaluate `lim_(xrarr(pi)/(2))(cosx)/(((pi)/(2)-x)).`

Answer» Put `(x-(pi)/(2))=y,` so that when `xto(pi)/(2)then tyto0.`
`thereforelim_(xto(pi)/(2))(cosx)/(((pi)/(2)-x))=lim_(yto0)(cos((pi)/(2)+y))/(-y)=lim_(yto0)((-siny)/(-y))=lim_(yto0)(siny)/(y)=1.`
Discussion
131.

Evaluate `lim_(xrarr(pi)/(6))((sqrt3sinx-cosx))/((x-(pi)/(6))).`

Answer» `lim_(x to(pi)/(6))(x-(pi)/(6))`
`=lim_(xto(pi)/(6))(2((sqrt3)/(2)sinx-1/2cosx))/((x-(pi)/(6)))=lim_(xto(pi)/(6))(2(sinxcos""(pi)/(6)-cosxsin""(pi)/(6)))/((x-(pi)/(6)))`
`=2lim_(x to (pi)/(6))(sin(x-(pi)/(6)))/((x-(pi)/(6)))=2lim_(yto0)(siny)/(y)=2" ""[""putting"(x-(pi)/(6))=y"]".`
Discussion
132.

If f , g , and h are functions having a cammon domain D and `h(x) lef(x) leg(x), x in D` and if `lim_(x to a)h(x)=lim_(x to a)g(x)=l" "then" "lim_(x to a)f(x)=l` The value of `lim_(x to 0)(|x|)/(sqrt(x^(4)+4x^(2)+7))-`A. 1B. 0C. `(1)/(2)`D. `-(1)/(2)`

Answer» Correct Answer - B
Discussion
133.

Evaluate `lim_(x to0)(xtanx)/((1-cosx))`

Answer» `lim_(x to 0)(x tanx)/((1-cosx))=lim_(xto0)(x tanx)/(cos x(1-cosx))`
`=lim_(x to 0)(x sin^(2)x)/(sinx cos x(1-cosx))`
`=lim_(x to 0)(x(1-cos^(2)x))/(sinx cosx(1-cosx))=lim_(x to 0)(x(1+cosx))/(sin x cosx)`
`=lim_(xto0)(x)/(sinx)lim_(xto0)((1+cosx))/(cosx)=(1xx(2)/(1))=2.`
Discussion
134.

If f , g , and h are functions having a cammon domain D and `h(x) lef(x) leg(x), x in D` and if `lim_(x to a)h(x)=lim_(x to a)g(x)=l" "then" "lim_(x to a)f(x)=l` `lim_(x to 0)x^(4)sin((1)/(3sqrtx))` is

Answer» Correct Answer - A
Discussion
135.

Evaluate the following limits: `lim_(x to0)(sqrt2-sqrt(1+cosx))/(2x+sin3x)`

Answer» Correct Answer - `(1)/(4sqrt2)`
`lim_(xto0)(sqrt2-sqrt(1+cosx))/(sin^(2)x)=lim_(xto0)(sqrt2-sqrt(2cos^(2)(x//2)))/((1-cos^(2)x))=lim_(xto0)(sqrt2(1-cos""(x)/(2)))/((1-cosx)(1+cosx))`
`=lim_(xto0)(sqrt2(1-cos""(x)/(2)))/((2sin^(2)""(x)/2)(1+cosx))=(1)/(sqrt2)lim_(xto0)((1-cos""(x)/(2)))/((1-cos^(2)""(x)/(2))(1+cosx))`
`=(1)/(sqrt2)lim_(xto0)(1)/((1+cos""(x)/(2))(1+cosx))=(1)/(sqrt2)xx(1)/((1+1)(1+1))=(1)/(4sqrt2).`
Discussion
136.

`lim_(x->1) ((x^4-3x^2+2)/(x^3-5x^2+3x+1))`

Answer» Correct Answer - `1/2`
Given limit `=lim_(xto1)((x-1)(x+1)(x^(2)-2))/((x-1)(x^(2)-4x-1))=lim_(xto1)((x+1)(x^(2)-2))/((x^(2)-4x-1))=1/2.`
Discussion
137.

Let `f(x)={{:(4x-5",",xle2),(x-a",",xgt2.):}` If `lim_(xrarr2)f(x)` exists then find the value of a.

Answer» Correct Answer - `a=-1`
`lim_(xto2^(-))f(x)=lim_(hto0)f(2+h)=lim_(hto0)(2+h-a)=(2-a).`
`lim_(xto2^(-))f(x)=lim_(hto0)f(2-h)=lim_(hto0){4(2-h)-5}=(8-5)=3.`
Since `lim_(xto2^(-))f(x)` exist, we must have `lim_(xto2^(+))f(x)=lim_(xto2^(-))f(x).`
`therefore2-a=3impliesa=(2-3)=-1.`
Discussion
138.

Let `f(x)={{:(4x-5",",xle2),(x-a",",xgt2.):}` If `lim_(xrarr2)f(x)` exists then find the value of a.

Answer» Correct Answer - `a=-1`
`lim_(xto2^(-))f(x)=lim_(hto0)f(2+h)=lim_(hto0)(2+h-a)=(2-a).`
`lim_(xto2^(-))f(x)=lim_(hto0)f(2-h)=lim_(hto0){4(2-h)-5}=(8-5)=3.`
Since `lim_(xto2^(-))f(x)` exist, we must have `lim_(xto2^(+))f(x)=lim_(xto2^(-))f(x).`
`therefore2-a=3impliesa=(2-3)=-1.`
Discussion
139.

`lim_(x->pi)(sin(pi-x)/(pi(pi-x)))`

Answer» Correct Answer - `(1)/(pi)`
Putting `(pi-x)=h, "we have" x to pi implies pi-xto0rarrhto0.`
`thereforelim_(xto pi)(sin(pi-x))/(pi(pi-x))=(1)/(pi)lim_(hto0)(sinh)/(h)=((1)/(pi)xx1)=(1)/(pi).`
Discussion
140.

Let `f(x){:{(2x+3",",xle0),(3(x+1)",",xgt0.):}` Find `(i) lim_(xrarr0)f(x)" "(ii)lim_(xrarr1)f(x)`

Answer» (I) We have
`lim_(xto0^(+))f(x)=lim_(hto0)3(0+h+1)=lim_(hto0)3(h+1)=3.`
`lim_(xto0^(-))f(x)=lim_(hto0)3(0-h+1)=lim_(hto0)3(-h+1)=3.`
`thereforelim_(xto0^(+))(x)=lim_(xto0^(-))f(x)=3.`
Hence, `lim_(xto0)f(x)=3.`
(ii) We have
`limf(x)lim_(hto0)f(1+h)=lim_(hto0)3(1+h+1)=lim_(hto0)3(2+h)=6.`
`lim_(xto1^(-))f(x)=lim_(hto0)f(1-h)=lim_(hto0)3(1-h+1)=lim_(hto0)3(2-h)=6.`
`thereforelim_(xto1^(+))f(x)=lim_(xto1)f(x)=6.`
Hence `lim_(xto1)f(x)=6.`
Discussion
141.

Evaluate the following limits: `lim_(xrarr0)((e^(bx)-e^(ax))/(x)),0ltaltb`

Answer» Correct Answer - `(b-a)`
Given limit `=lim_(xto0){((e^(bt)-1)-(e^(ax)-1))/(x)}=lim_(bxto0){((e^(bx)-1)/(bx)).b}-lim_(axto0){((e^(ax)-1)/(ax)).a}`
`=(1xxb)-(1xxa)=(b-a).`
Discussion
142.

Evaluate the following limits: `lim_(xrarr0)(cos ax-cos bx)/(cos cx-1)`

Answer» Correct Answer - `((a^(2)-b^(2)))/(c^(2))`
Given limit `=lim_(xto0)(-2sin((a+b)/(2))x.sin((a-b)/(2))x)/(-2sin^(2)((cx)/(2)))`
`((sin((a+b)/(2))x)/(((a+b)/(2)))xx((a+b)/(2))x.(sin((a-b)/(2))x)/(((a-b)/(2))x)xx((a-b)/(2))x)/({(sin((cx)/(2)))/(((cx)/(2)))}^(2))`
`={((a+b)/(2))((a-b)/(2))xx(4)/(c^(2))}.lim_(xto0)(sin((a+b)/(2))x)/(((a+b)/(2))).lim_(xto0)(sin((a-b)/(2))x)/(((a-b)/(2))x)`
`={((a^(2)-b^(2)))/(x^(2))xx1xx1}=((a^(2)-b^(2)))/(c^(2)).`
Discussion
143.

Evaluate the following limits: `lim_(xrarr0)(sinax+bx)/(ax+sinbx),where a, b, a+bne0`

Answer» Correct Answer - 1
`lim_(xto0)(sina+bc)/(ax+sin bx)=lim_(xto0){((sinax)/(x)+b)/(a+(sinbx)/(x))}" "["dividing num. and denom. by x"]`
`=lim_(xto0)({a((sinax)/(ax))+b})/({a+b((sinbx)/(bx))})=((axx1)+b)/(a+(bxx1))=(a+b)/(a+b)=1.`
Discussion
144.

Find `(dy)/(dx)` at `x=-1` ,when `(siny)^(sin((pi/2)x))+sqrt3/2 sec^(-1)(2x)+2^x tan(ln(x+2))=0`

Answer» Correct Answer - `3/(pisqrt(pi^(2)-3)) `
Here, `(siny)^(sin.pi/2x) +sqrt3/2 sec^(-1)(2x)+2^(x) tan {log (x+2)}=0`
On differentiating both sides, we get
`(sin y)^(sin.pi/2 x)*log (sin y)*cos.pi/2 x*pi/2`
`" "+(sin. pi/2x)(sin y)^((sin .pi/2 x)-1)*cos y *(dy)/(dx) `
`" "+sqrt3/2*2/((2|x|)sqrt(4x^(2)-1))+(2^(x)*sec^(2){log (x+2)})/((x+2))`
`" "+ 2^(x) log 2* tan {log (x+2)}=0`
Putting `(x=- 1, y = - sqrt3/pi)`, we get
`(dy)/(dx)=(-sqrt3/pi)^(2)/sqrt(1-(sqrt(3)/pi)^(2))=3/(pisqrt(pi^(2)-3))`
Discussion
145.

Let `p=(lim)_(xvec0+)(1+tan^2sqrt(x))^(1//2x)`then `logp`is equal to:(1) 2(2) 1(3) `1/2`(4)`1/4`A. 2B. 1C. `1/2`D. `1/4`

Answer» Correct Answer - C
Given, ` p = underset( x to 0^(+)) lim (1+ tan^(2) sqrtx)^(1/(2x) ) ( 1^( infty) " form") `
` = e^(underset( x to 0^(+)) lim (tan^(2)sqrtx)/(2x)) = e^(1/2 underset( x to 0^(+)) lim ((tan sqrtx)/(sqrtx))^(2))=e^(1/2) `
` :. " " log p = log e^(1/2) = 1/2 `
Discussion
146.

Evaluate the following limits: `lim_(xrarr0)((sqrt(1+x+x^(2))-1)/(x))`

Answer» Correct Answer - `1/2`
Discussion
147.

Evaluate the following limits: `lim_(xrarr0)((sqrt(2-x)-sqrt(2+x))/(x))`

Answer» Correct Answer - `(-1)/(sqrt2)`
Discussion
148.

Evaluate the following limits: `lim_(xrarr5)((x^(2)-25)/(x-5))`

Answer» Correct Answer - 10
Discussion
149.

Evaluate the following limits: `lim_(xrarr-2)((x^(3)+8)/(x+2))`

Answer» Correct Answer - 12
Discussion
150.

Evaluate the following limits: `lim_(xrarr3)((x^(2)-4x+3)/(x^(2)-2x-3))`

Answer» Correct Answer - `1/2`
Discussion