Explore topic-wise InterviewSolutions in .

This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.

251.

`lim_(xrarr0) x^2sin.(pi)/(x)`, isA. 1B. 0C. non-existantD. `oo`

Answer» Correct Answer - B
Discussion
252.

The value of `lim_(xrarr0){tan((pi)/(4)+x)}^(1//pi)`, isA. eB. `e^2`C. `(2)/(e)`D. `(1)/(e^2)`

Answer» Correct Answer - B
We have,
`lim_(xto 0) {tan ((pi)/(4)+x)}^((1)/(x))=lim_(xto0) ((1+tanx)/(1-tanx))^(1//x)`
` =lim_(xto0) (1+(2tanx)/(1-tanx))^(1//x)=e ^(lim_(xto 0) (2tanx)/(1-tanx ).(1)/(x))`
`=e^(lim_(xto0) (2)/(1-tanx)xx(tanx)/(x))=e^2` .
Discussion
253.

Evaluate `lim_(xto1) (x^(2)+xlog_(e)x-log_(e)x-1)/((x^(2))-1)`

Answer» `underset(xto1)lim(x^(2)+xlog_(e)x-log_(e)x-1)/((x^(2))-1)" "(0/0"from")`
`=underset(xto1)lim((x-1)(log_(e)x+x+1))/((x+1)(x-1))" "(0/0"from")`
`=underset(xto1)lim(log_(e)x+x+1)/(x+1)`
`=(log_(e)1+1+1)/(1+1)=(0+2)/(2)=1`
Discussion
254.

Find the following limits: `(i) lim_(xto0) (1-x)^((1)/(x))" "(ii) lim_(xto1) (1+log_(e)x)^((1)/(log_(e)x))` `(iii)lim_(xto0) (1+sinx)^((1)/(x))`

Answer» `(i) underset(xto)lim(1-x)^((1)/(x))=(underset(xto0)lim(1-x)^((1)/(-x)))^(-1)=e^(-1)`
(ii) Since, `underset(xto1)log_(e)x=0, underset(xto1)lim(1+log_(e)x)^((1)/(log_(e)x))=1`
`(iii)underset(xto0)lim (1+sinx)^((1)/(x))=underset(xto0)lim((1+sinx)^((1)/(sinx)))^((sinx)/(x))`
`=(underset(xto0)lim(1+sinx)^((1)/(sinx)))^(underset(xto0)lim^((sinx)/(x)))`
`=e^(1)=e`
Discussion
255.

Evaluate `lim_(xto0^(+)) (1)/(x)cos^(-1)((sinx)/(x))`.

Answer» `underset(xto0^(+))lim(1)/(x)cos^(-1)((sinx)/(x))`
`=underset(xto0^(+))lim(sin^(-1)(sqrt(1-(sin^(2)x)/(x^(2)))))/(x)`
`=underset(xto0^(+))lim(sin^(-1)(sqrt(1-(sin^(2)x)/(x^(2))))sqrt(1-(sin^(2)x)/(x^(2))))/sqrt(1-(sin^(2)x)/(x^(2)))`
`=underset(xto0^(+))lim(sqrt(x^(2)-sin^(2)x))/(x^(2))`
`=underset(xto0^(+))limsqrt((x+sinx)/(x).(x-sinx)/(x^(3)))`
`=sqrt(2)underset(xto0^(+))limsqrt((x-(x-(x^(3))/(3!)))/(x^(3)))`
`=sqrt(2)sqrt((1)/(6))`
`=sqrt((1)/(3))`
Discussion
256.

The value of `lim_(xrarr0) (a^x-b^x)/(x)` ,isA. `log(a//b)`B. `log(b//a)`C. `log(ab)`D. `-log(ab)`

Answer» Correct Answer - A
Discussion
257.

Let the sequence `ltb_(n)gt` of real numbers satisfy the recurrence relation `b_(n+1)=1/3(2b_(n)+(125)/(b_(n)^(2))),b_(n)ne0.` Then find `lim_(ntoo0) b_(n).`

Answer» Correct Answer - 5
Let `underset(ntooo)limb_(n)=b`
Now `b_(n+1)=1/3(2b_(n)+(125)/(b_(n)^(2)))`
`implies" "underset(ntooo)limb_(n+1)=1/3(2underset(ntooo)limb_(n)+(125)/(underset(ntooo)limb_(n)^(2)))`
`implies" "b=1/3(2b+(125)/(b^(2)))" "( :.underset(ntooo)limb_(n)=underset(ntooo)limb_(n+1)=b)`
`implies" "b/3=(125)/(3b^(2))`
`implies" "b^(3)=125`
`implies" "b=5`
Discussion
258.

Evaluate`("lim")_(nvecoo){cos(x/2)cos(x/4)cos(x/8) cos(x/(2^n))}`A. 1B. `(sinx)/(x)`C. `(x)/(sinx)`D. none of these

Answer» Correct Answer - B
Discussion
259.

The value of `lim_(xrarr0) (e^x-(x+x))/(x^2)`,is

Answer» Correct Answer - B
Discussion
260.

Evaluate `lim_(xto1) "sec" (pi)/(2^(x)).log_(e)x`.

Answer» We have, `underset(xto1)lim(log_(e)x)/("cos"(pi)/(2^(x)))=underset(xto1)lim(log(1+(x-1)))/(underset(xto1)lim(sin((pi)/(2)-(pi)/(2^(x))))/(((pi)/(2)-(pi)/(2^(x))))).underset(xto1)lim(x-1)/((pi)/(2)-(pi)/(2^(x)))`
`=underset(xto1)lim(x-1)/(pi((2^(x-1)-1)/(2^(x))))`
`=underset(xto1)lim(2^(x))/(pi).(x-1)/(2^(x-1)-1)`
`=(2)/(pilog2)`
Discussion
261.

The value of `lim_(xrarr0) (int_(0)^(x)tdt)/(xtan (x+pi))` is equal to

Answer» Correct Answer - C
Discussion
262.

`lim_(xrarr1)(1-x)tan((pix)/2)` is equal toA. `pi//2`B. `pi+2`C. `2//pi`D. none of these

Answer» Correct Answer - C
Discussion
263.

If`f(x)={sinx ,x!=npi,n in I2,ot h e r w i s e``g(x)={x^2+1,x!=0,4,x=0 5,x=2`then `("lim")_(xvec0)g{f(x)}i s=`A. 1B. 5C. 6D. 7

Answer» Correct Answer - A
Discussion
264.

`lim_(xrarr1) (1+cos pix)cot^2pi x` is equal toA. 1B. `-1`C. `1//2`D. `-1//2`

Answer» Correct Answer - C
Discussion
265.

Let the sequence ``real numbers satisfies the recurrence relation `b_(n+1)=1/3(2b_n+(125)/(b n2)),b_n!=0.`Then find the`("lim")_(nvecoo)b_ndot`

Answer» Correct Answer - C
Let `lim_(nto oo)b_(n)=b`. Then,
` b_(n+1)=(1)/(3)(2b_n+(125))/(b_(n^(2))), b_(n) ne 0`, then, `lim_(nto oo) b_n=` then, `lim_(nto oo) b_n=`
`rArr lim_(nto oo) b_(n+1)=(1)/(3) {2 lim_(nto oo) b_(n)+(125)/(lim_(nto oo) b_n^(2))}`
` rArr b=(1)/(3) {2b +(125)/(b^2)}`
` rArr (b)/(3)=(125)/(3b^2)rArr b^3=125rArr b=5`
Discussion
266.

Evaluate :`("lim")_(xvecoo)(a x^2+b x+c)/(dx^2+e x+f)dot`A. `(a)/(a)`B. `(d)/(a)`C. `(b)/(e)`D. `(c)/(f)`

Answer» Correct Answer - A
Here , the expression assumes the form `(oo)/(oo)`.
We notice that the higest power of x in both the numerator and denominator is 2. So, we divide each each term in both the numerator and denominator by `x^2`.
Discussion
267.

`lim_(xrarr1)(sin(e^(x-1)-1))/(log x)` is equal toA. 1B. 0C. eD. `e^-1`

Answer» Correct Answer - A
Discussion
268.

Evaluate, `underset(xto1)"lim"(x^(4)-1)/(x-1)= underset(xtok)"lim"(x^(3)-k^(3))/(x^(2)-k^(2))` , then find the value of k.A. `(4)/(3)`B. `(8)/(3)`C. `(2)/(3)`D. none of these

Answer» Correct Answer - B
We have,
`lim_(xto1)(x^4-1)/(x-1)=lim_(xto1)(x^4-1^4)/(x-1)=4(1)^(4-1)=4`
and `lim_(x to k) (x^3-k^3)/(x^2-k^2)`
` =lim_(x to k) (x^3-k^3)/(x-k)xx(x-k)/(x^2-k^2)`
`lim_(x to k) (x^3-k^3)/(x-k)div (x^2-k^2)/(x-k)`
` lim_(xto k) (x^3-k^3)/(x-k)div lim_(x to k)(x^3-k^3)/(x -k) = (3k^2)/(2k)=(3)/(2)k`
`therefore lim_(x to 1) (x^4-1)/(x-1)=lim_(x tok)(x^3-k^3)/(x^2-k^2)rArr 4=(3k)/(2)rArrk=(8)/(3)`.
Discussion
269.

The value of `lim_(xto0) (1+sinx-cosx+log(1-x))/(x^(3))` isA. 1B. -1C. 2D. -2

Answer» Correct Answer - B
`underset(xto0)lim(1+sinx-cosx+log(1-x))/(x^(3))`
`=underset(xto0)lim(1+(x-(x^(3))/(3!)+...)-(1-(x^(2))/(2!)+(x^(4))/(4!)-...)+(-x-(x^(2))/(2)-(x^(3))/(3)-...))/(x^(3))`
`=-(1)/(3!)-(1)/(3)=-(1)/(2)`
Discussion
270.

The value of `lim_(xrarr1) (2-x)^(tan.(pix)/(2))` is equal toA. `e^(-2//pi)`B. `e^(1//pi)`C. `e^(2//pi)`D. `e^(-1//pi)`

Answer» Correct Answer - C
Discussion
271.

If `(lim)_(x->-a)(x^9+a^9)/(x+a)9, `find the real value of `adot`A. `+-1`B. `+-3`C. `+-2`D. none of these

Answer» Correct Answer - A
We have, `lim_(x to pi//4)(x^9+a^9)/(x+a)=9`
`rArr lim_(x to -a)(x^9-(-a)^9)/(x-(-a))=9`
`rArr 9(-a)^(9-1)=9rArr 9a^8=9rArr a ^8=1 rArr a=+-1`
Discussion
272.

If `lim_(xrarr1)(x+x^2+x^3+....+x^n-n)/(x-1)=5050` , then `n=`A. 10B. 100C. 150D. none of these

Answer» Correct Answer - B
We have,
`lim_(x to oo) (x+x^2+x^3+....+x^n-n)/(x-1)=5050`
`rArr (n(n+1))/(2)=5050 ["See illustration" 3]`
`rArr n=100`
Discussion
273.

the value of `lim_(x->0)(cos(sinx)-cosx)/x^4` is equal to:A. `(1)/(5)`B. `(1)/(6)`C. `(1)/(4)`D. `(1//2)`

Answer» Correct Answer - B
Discussion
274.

`lim_(x-> pi/4) (4sqrt2-(cosx+sinx)^5)/(1-sin2x)`A. `5sqrt(2)`B. `3sqrt(2)`C. `sqrt(2)`D. none of these

Answer» Correct Answer - A
`lim_(xto pi//4)(4sqrt(2)-(cosx+sinx)^5)/(1-sin2x)`
`=lim_(x+pi//4)({(cos x +sin x)^2}^(5//2)-(2)^(5//2))/((1+sin2x)-2)`
`=lim_(x to pi//4)((1+sin2x)^(5//2)-2^(5//2))/((1+sin2x)-2)`
`=lim_(y+2)(y^(5//2)-2^(5//2))/(y-2),"where " y=1 +sin 2x`
`=(5)/(2)xx2^(5//2-1)=5sqrt(2)`
Discussion
275.

Evaluate `lim_(xto(pi)/(4)) (1-sin2x)/(1+cos4x).`

Answer» Correct Answer - `1//4`
`underset(xto(pi)/(4))lim(1-sin2x)/(1+cos4x)" "`(0/0 form)
`=underset(xto(pi)/(4))lim(sinx-cosx)^(2)/(2cos^(2)2x)" "`(0/0 form)
`=underset(xto(pi)/(4))lim(sinx-cosx)^(2)/((cos^(2)x-sin^(2)x)^(2))" "`(0/0 form)
`=underset(xto(pi)/(4))lim(1)/(2(cosx+sinx)^(2))=1/4`
Discussion
276.

The value of ` lim_(xrarr 0) (1-cos(1-cos x))/(x^4)` is equal toA. `(1)(8)`B. `(1)/(2)`C. `(1)/(4)`D. none of these

Answer» Correct Answer - A
Discussion
277.

`lim_(xrarr0)(4^x-1)/(3^x-1)` equalsA. `log_3 4`B. `log_4 3`C. `log_e 4`D. `log_3 4`

Answer» Correct Answer - A
We have,
`lim_(xto0)(4^x-1)/(3^x-1)=lim_(xto0) (4x-1)/(x)xx(x)/(3^x-1)=(log_e4)/(log_e3)=log_3 4 `
Discussion
278.

The value of `lim_(x->pi/2)tan^2x( sqrt[2sin^2x+3sinx+4]- sqrt[sin^x+6sinx+2])` is equal toA. `(1)/(10)`B. ` (1)/(11)`C. `( 1)/(120`D. ` (1)/(8)`

Answer» Correct Answer - C
Discussion
279.

`lim_(x->pi/4)(2sqrt(2)-(cosx+sinx)^3)/(1-sin2x)=`A. `(3)/(sqrt(2))`B. `(sqrt(2))/(3)`C. `(1)/(sqrt(2))`D. `sqrt(2)`

Answer» Correct Answer - A
Discussion
280.

`lim_(xto0) {(1)/(x3sqrt(8+x)-(1)/(2x))}` is equal toA. `(1)/(12)`B. `(-4)/(3)`C. `(-16)/(3)`D. `(-1)/(48)`

Answer» Correct Answer - D
Discussion
281.

`lim_(xrarr0)(cos (sinx)-1)/(x^2)=`A. 1B. -1C. `1//2`D. `-1//2`

Answer» Correct Answer - D
We have,
` lim_(xto0) (cos(sinx)-1)/(x^2)`
`=-(1)/(2)xx1[because lim_(xto0) (1-cos x) /x^2=(1)/(2)]`.
Discussion
282.

`lim_(x to 0) ((1-cos2x)(3+cosx))/(xtan4x)` is equal to

Answer» Correct Answer - C
`underset(xto0)lim((1-cos2x)(3+cosx))/(xtan4x)=underset(xto0)lim((2sin^(2)x)(3+cosx))/(x((tan4x)/(4x))xx4x)`
`=underset(xto0)lim(2sin^(2)x(3+cosx))/(4x^(2))`
`=(2)/(4)(3+1)=2`
Discussion
283.

`lim_(x rarr -a) (x^(n)+a^(n))/(x+a)` (where n is an odd natural number)

Answer» Correct Answer - `n(-a)^(n-1)`
Discussion
284.

Evaluate: `lim_(x rarr 1) (f(x)-f(1))/(x-1), "where" f(x) = x^(2)-2x`.A. -1B. 0C. 1D. 2

Answer» Correct Answer - B
Calculate f(1) substitute f(x) and f(1), factorise numerator, then cancel the common factor and substitute x = 1.
Discussion
285.

Evaluate: `lim_(x rarr 0) (sqrt(6+x)-sqrt(6-x))/(sqrt(8+x)-sqrt(8-x))`.A. `sqrt(3)`B. `(-5sqrt(3))/(3)`C. `(4sqrt(3))/(3)`D. `(2sqrt(3))/(3)`

Answer» Correct Answer - D
Rationalize both numerator and denominator.
Discussion
286.

If `lim_(x rarr a) (x^(7)-a^(7))/(x-a)=7`, then find the number of possible real values of a.

Answer» Correct Answer - C
(i) Use the formula, `underset(x rarr a)("lim")(x^(m)-a^(m))/(x^(n)-a^(n))=(m)/(n)a^(m-n)`
(ii) `7a^(6) = 7`.
(iii) Now, find the number of possible real values of a.
Discussion
287.

Evaluate: `lim_(x rarr 5) sqrt(25-x^(2))`.

Answer» Correct Answer - Limit does not exist
Discussion
288.

Evaluate `lim_(x rarr 3) ((3 - sqrt(6+x)))/(x-3)`.

Answer» When `x=3, ((3-sqrt(6+3)))/(3-3)=(0)/(0)`, which is an indeterminate form. As f(x) is irrational, we multiply the numerator and denominator with the rationalizing factor of f(x).
`underset(x rarr 3)("lim")((3-sqrt(6+x)))/(x-3)=underset(x rarr 3)("lim")((3-sqrt(6+x)))/(x-3)xx((3+sqrt(6+x)))/((3+sqrt(6+x)))=underset(x rarr 3)("lim")(9-(6+x))/((x-3)(3+sqrt(6+x)))`
`" "underset(x rarr 3)("lim")(3-x)/((x-3)(3+sqrt(6+x)))=underset(x rarr 3)("lim")(-1)/((3+sqrt(6+x)))=(-1)/((3+sqrt(6+3)))`
`" "underset(x rarr 3)("lim")(-1)/((3+3))=-(1)/(6)`.
Discussion
289.

`lim_(xto0) (sin(picos^(2)x))/(x^(2))` is equal to

Answer» Correct Answer - D
`underset(xto0)lim(sin(picos^(2)x))/(x^(2))=underset(xto0)lim(sin(pi-pisin^(2)x))/(x^(2))`
`=underset(xto0)lim(sin(pisin^(2)x))/(pisin^(2)x)xx(pisin^(2)x)/(x^(2))=pi`
Discussion
290.

`lim_(x->(pi)/6)(3sinx-sqrt(3)cosx)/(6x-pi)`A. `sqrt(3)`B. `(1)/(sqrt(3))`C. `-sqrt(3)`D. `-(1)/(sqrt(3))`

Answer» Correct Answer - B
We have,
` lim_(xto pi//6) (3sinx-sqrt(3)cosx)/(6x-pi)`
` =lim_(xtopi//6) (2sqrt(3)(sqrt(3)/2sinx-(1)/(2)cosx ))/(6(x-(pi)/(6)))`
`lim_(xtopi//6)(1)/(sqrt(3)) sin(x-(pi)/(6))/((x-(pi)/6))=(1)/(sqrt(3))`
Discussion
291.

`lim_(xto0) (sin(picos^2x))/(x^2)` equalsA. `-pi`B. ` pi`C. `pi//2`D. `1`

Answer» Correct Answer - B
`lim_(xto0)((sinpicos^2x))/(x^(2))((0)/(0)"from")`
` rArr lim_(ato0) (sin(pi-pisin^2x))/(x^2)`
` rArr lim_(ato0) (sin(pisin^2x))/(pisin^2x) pi ((sinx)/(x))^2=1xxpi(1)^2=pi`
Discussion
292.

`lim_(xrarr0) (sin(picos^2x))/(x^2)` equalsA. `-pi`B. `pi`C. `(pi)/(2)`D. `1`

Answer» Correct Answer - B
We have ,
`lim_(xto0) (sin(picos^2x))/(x^2)`
` =lim_(xto0) (sin(pi-picos^2x))/(x^2)`
` =lim_(xto0) (sin(pisin^2x))/(pisin^2x)xx(pisin^2x)/(x^2)=1xxpixx(1)^2=pi`
Discussion
293.

Evaluate the following limit : \(\lim\limits_{\text x \to -1}\)(4x2 + 2)lim(x→-1)(4x2 + 2)

Answer»

Given limit \(\Rightarrow\lim\limits_{\text x \to -1} \)(4x2 + 2)

Putting the value of limits directly, we have

\(\Rightarrow\)(4(-1)2 + 2)

\(\Rightarrow\) (4(1) + 2)

\(\Rightarrow\) 6

Hence the value of the given limit is 6.
Discussion
294.

`lim_(x rarr 1) (root(5)(x)-1)/(root(4)(x)-1)=`_______.

Answer» Correct Answer - `(4)/(5)`
Discussion
295.

The value of `lim_(xrarr0) (sinx-x+(x^3)/(6))/(x^5)`, is

Answer» Correct Answer - D
Discussion
296.

Evaluate: `lim_underset(x rarr 0) (sqrt(1+x+x^(2)+x^(3))-1)/(x)`.

Answer» Correct Answer - `(1)/(2)`
Discussion
297.

Evaluate `lim_(x rarr 0) [(x)/(1 - sqrt(1-x))]`.

Answer» When x = 0, the given function assumes the form `(0)/(1- sqrt(1-0))=(0)/(0)`, which is an indeterminate form.
Since g(x), the denominator is an irrational form, we multiply the numerator and denominator with the rationalizing factor of g(x).
`underset(x rarr 0)("lim")[(x)/(1 - sqrt(1-x))]=underset(x rarr 0)("lim")(x)/(1 - sqrt(1-x)) xx ((1 + sqrt(1-x)))/((1 + sqrt(1-x)))`
`" "= underset(x rarr 0)("lim")(x(1 + sqrt(1-x)))/(1 - 1 + x)`
`" "= underset(x rarr 0)("lim")(x(1 + sqrt(1-x)))/(x)`
`" "= underset(x rarr 0)("lim")(1 + sqrt(1-x))=1 + sqrt(1) = 2`.
Discussion
298.

For some real number k, the value of `lim_(x rarr -k) (x^(5)+k^(5))/(x+k)` can be____.A. 50B. 60C. 70D. 80

Answer» Correct Answer - D
Use the formula, `underset(x rarr a)("lim")(x^(n)-a^(n))/(x-a)=na^(n-1)`
Discussion
299.

The value of `lim_(xto pi//2)(cot x-cosx)/(pi-2x)^3` isA. `(1)/(2)`B. `(1)/(4)`C. `(1)/(8)`D. `(1)/(16)`

Answer» Correct Answer - D
We have,
` lim_(xtopi//2)(cotx-cosx)/((pi-2x)^3)`
`= lim_(xtopi//2)(tan((pi)/(2)-x)-sin((pi)/2-x))/(8((pi)/(2)-x)^3)=(1)/(8)xx(1)/(2)=(1)/(16)`
Discussion
300.

Evaluate `lim_(xto0) sin(picos^(2)x)/(x^(2)).`

Answer» `underset(xto0)limsin(picos^(2)x)/(x^(2))=underset(xto0)limsin(pi-picos^(2)x)/(x^(2))`
`=underset(xto0)limsin(pisin^(2)x)/(pisin^(2)x)xx(pisin^(2)x)/(x^(2))=pi`
Discussion