Mathematics Previous year questions of Limits for NDA

Please Wait... While Loading Full Video

Previous year questions of Limits for NDA

Q 2743491343

What is `lim_( x→ 0) (e^x - (1+x))/x^2` equal to ?
NDA Paper 1 2017

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

`lim _(x-> 0) (e^x - (1+x))/(x^2)`

`(0/0)` so by L'Hospital

`lim_(x-> 0) (e^x -1)/(2x)`

`0/0` form so by L'Hospital

`lim_(x-> 0) (e^x )/2 =1/2`

Correct Answer is `=>` (B) `1/2`

Q 2703691548

If `F(x) = sqrt(9-x^2)` then what is `lim_( x→ 1) (F(x) - F(1))/(x-1)` equal to?
NDA Paper 1 2017

(A)

`-1/(4 sqrt2)`

(B)

`1/8`

(C)

`-1/(2 sqrt2)`

(D)

`1/(2 sqrt2)`

Solution:

`F(x) = sqrt(9-x^2)`

`lim_(x-> 1) (F(x) -F(1))/(x-1) = F'(1)`

`[((-2x))/(2 sqrt(9-x^2))]_(x=1)=(-1) /(2 sqrt 2)`

Correct Answer is `=>` (C) `-1/(2 sqrt2)`

Q 2703491348

Consider the following statements

1. If `lim_(x-> a) f(x)` and `lim_(x-> a) g(x)` both exist, then `lim_(x-> a) {f(x) g(x)}` exists.

2. If `lim_(x-> a) f(x)g(x)` exists , then both `lim_(x-> a) f(x) and lim_(x-> a) g(x)` must exists.

Which of the above statements is/are correct?
NDA Paper 1 2017

(A)

1 only

(B)

2 only

(C)

Both 1 and 2

(D)

Neither 1 nor 2

Solution:

We are not supposed to prove it.

First is obvious and we know it from theory as well.

For 2nd let `f(x)` be `x-a` and `g(x)` be `1/(x-a)`

Then

`lim_(x-> a) f(x) g(x) = lim_(x-> a) ((x-a))/((x-a))=1`

`lim_(x-> a) f(x)=0` and `lim_(x-> a) g(x) =lim_(x-> a) 1/(x-a)` which doesn't exist.

Correct Answer is `=>` (A) 1 only

Q 2116512479

Consider the function
` f(x) = (a^([x] + x ) - 1 )/([x] + x )`
where [ . ] denotes the greatest integer function.

What is `lim_(x->0^(+)) f(x)` equal to

NDA Paper 1 2016

(A)

`1`

(B)

ln a

(C)

`1- a^(-1)`

(D)

Limit does not exist

Solution:

Given, `f(x) = (a^([x] + x ) - 1 )/([x] + x )`

Now, RHL `= lim_(x->0^(+)) (a^([x] + x ) - 1 )/([x] + x )`

As, `x -> 0^(+)` (i.e. approaches 0 from the right), we get
`[x] =0`

`:. lim_(x->0^(+)) f(x) = lim_(x->0^(+)) (a^(x) -1)/x`

` = lim_(h -> 0) (a^(0 + h ) -1 )/(0+h) = lim_(h -> 0) (a^(h) - 1)/h = ln a`

Correct Answer is `=>` (B) ln a

Q 2176612576

Consider the function
` f'(x) = (a^([x] + x ) - 1 )/([x] + x )`
where [ . ] denotes the greatest integer function.

What is `lim_(x -> 0^(-)) f(x)` equal to?
NDA Paper 1 2016

(A)

`0`

(B)

In a

(C)

`1-a^(-1)`

(D)

Limit does not exist

Solution:

` lim_(x->0^(-)) f(x) = lim_(x->0^(-)) (a^([x] + x ) - 1 )/([x] + x )`

As `x -> 0^(-)` (i.e. approaches 0 from left), we get `[x] = -1`

`:. lim_(x->0^(-)) f(x) = lim_(x->0^(-)) (a^(-1+x) -1)/(-1 +x)`

` = lim_(h -> 0^(-)) (a^(-1+(0 - h)) -1)/(-1 +(0-h)) = (a^(-1) -1 )/-1 = 1 - a^(-1)`

Correct Answer is `=>` (C) `1-a^(-1)`

Q 2137101082

If `lim_(x -> 0) phi (x) = a^(2)` , where `a != 0`, then what is `lim_(x -> 0) phi (x/a)`
equal to?
NDA Paper 1 2016

(A)

`a^(2)`

(B)

`a^(-2)`

(C)

`-a^(2)`

(D)

`-a`

Solution:

Given `lim_(x -> 0) phi (x) = a^(2)`

To find `lim _(x/a -> 0) phi (x/a) = a^2`

`:. lim _(x -> 0) phi (x/a) = a^(2)`

Correct Answer is `=>` (A) `a^(2)`

Q 2167101085

What is `lim_(x -> 0) e^(-1//x^(2))` equal to?
NDA Paper 1 2016

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

Limit does not exist

Solution:

`lim_(x -> 0) e^(-1//x^(2)) = lim_(h -> 0) e^(-1//(0 + h)^(2)) = 0`

Correct Answer is `=>` (A) `0`

Q 1608480308

If `G(x) = sqrt (25- x^2)` , then what is `lim_(x -> 1) ( G(x) -G(1) )/(x-1)`

equal to?
NDA Paper 1 2015

(A)

` - 1/(2 sqrt(6))`

(B)

`1/5`

(C)

` - 1/ sqrt(6)`

(D)

` 1/ sqrt(6)`

Solution:

We have , `G(x) = sqrt (25- x^2)`

To find : `lim_(x -> 1) ( G(x) -G(1) )/(x-1)`

Clearly, `lim_(x -> 1) ( G(x) -G(1) )/(x-1) = G'(1)` .......(1)

Now, let us first find `G'(x)`, which is given by

`G' (x) = 1/2 xx 1/ sqrt(25 - x^2) (-2x)`

`= G' (x) = (-x)/ sqrt(25 - x^2)`

` = > G' (1) = (-1)/sqrt(24) = (-1 )/(2sqrt(6))`

Correct Answer is `=>` (A) ` - 1/(2 sqrt(6))`

Q 2231278122

If `f(x) = sqrt(25- x^2)` , then what is `lim_(x -> 1) (f(x) - f (1) )/(x-1)` equal
to?
NDA Paper 1 2015

(A)

`1/5`

(B)

`1 /sqrt(24)`

(C)

` sqrt(24)`

(D)

`1 /-sqrt(24)`

Solution:

Consider `lim_(x -> 1) (f(x) - f (1) )/(x-1)`

` = lim_(x -> 1) (f'(x))/1 = lim_(x -> 1) (1(-2x))/(2 sqrt(25 - x^2))`

` = - 1/sqrt(24)`[by L'Hospital's rule]

Correct Answer is `=>` (D) `1 /-sqrt(24)`

Q 2251178924

If `f(x) = (sin (e^(x-2) -1)) / (ln (x-1))` ,then `lim_(x -> 2) f(x)` is equal to
NDA Paper 1 2015

(A)

`-2`

(B)

`-1`

(C)

`0`

(D)

`1`

Solution:

Consider, `lim_(x -> 2) (sin (e^(x-2) -1)) / (ln (x-1))`

` = lim_(x -> 2) (cos (e^(x-2) -1) . e^(x-2) ) / (1 /(x-1))`

` = (1 xx1) /1 =1`

Correct Answer is `=>` (D) `1`

Q 1619834710

Given that
` lim_(x-> oo) ( (2 + x^2)/(1 + x) - Ax - B) = 3`.

What is the value of `A?`
NDA Paper 1 2015

(A)

`-1`

(B)

`1`

(C)

`2`

(D)

`3`

Solution:

we have

` lim_(x-> oo) ( (2 + x^2)/(1 + x) - Ax - B) = 3`

` => lim_(x-> oo) ( (2 + x^2 -(Ax + B) (1 + x) ) /(1 + x) ) = 3`

` => lim_(x-> oo) ( (2 + x^2 -(Ax + Ax^2 + B + Bx)) /(1 + x) ) = 3`

` => lim_(x-> oo) ( (2 + x^2 - Ax - Ax^2 - B - Bx) /(1 + x) ) = 3`

` => lim_(x-> oo) ( (2x - A - 2 Ax - B )/1 ) = 3`

` => lim_(x-> oo) [x(2 - 2A)- (A + B)]= 3`

For limit to be exist and equal to `3`, we should have

`2 - 2A = 0` and `A + B = - 3`

`=> A=1`

Correct Answer is `=>` (B) `1`

Q 1649834713

Given that
` lim_(x-> oo) ( (2 + x^2)/(1 + x) - Ax - B) = 3`.

What is the value of `B?`
NDA Paper 1 2015

(A)

`-1`

(B)

`3`

(C)

`-4`

(D)

`-3`

Correct Answer is `=>` (C) `-4`

Q 1639167012

Consider the function `f(x) = {tt ( (x^2 - 5 , x <= 3) ,( sqrt(x + 13) , x > 3) )`

What is `lim_(x->3) f(x)` equal to?
NDA Paper 1 2014

(A)

`2`

(B)

`4`

(C)

`5`

(D)

`13`

Solution:

We have, `f(x) = tt { ( (x^2 - 5 , x <= 3) ,( sqrt(x + 13) , x > 3) )`

To find `lim_(x-> 3) f(x),`

`LHL = lim_(x-> 3^-) f(x)`

` = lim_(x-> 3^-) (x^2 - 5) `

` = lim_(x-> (3-h)) [(3-h)^2 - 5]`

` = lim_(h -> 0) (9 - 6h + h^2 -5) = 4`

`RHL = lim_(x-> 3^+) ( sqrt(x + 13))`

` = lim_(x-> (3+h)) ( sqrt(3 + h + 13))`

` = lim_(h -> 0) ( sqrt(16 + h)) = 4`

` ∵ lim_(x-> 3^-) f(x) = lim_(x-> 3^+) f(x) = 4`

`:. lim_(x-> 3) f(x) =4`

Correct Answer is `=>` (B) `4`

Q 1649867713

What is `lim_(x -> 0) ( log_5 (1 + x))/x` equal to.?
NDA Paper 1 2014

(A)

`1`

(B)

`log_5 e`

(C)

`log_e 5`

(D)

`5`

Solution:

`lim_(x -> 0) ( log_5 (1 + x))/x`

` = lim_(x -> 0) ( log_e (1+x))/(log_e 5x) `

` = 1/ (log_e 5) lim_(x -> 0) ( log_e (1+x))/x`

` = log_5 e ( ∵ lim_(x -> 0) ( log_e (1+x))/x = 1 ` and ` log_a b = 1/ ( log_b a))`

Correct Answer is `=>` (B) `log_5 e`

Q 1679867716

What is `lim_(x -> 0) (5^x -1)/x` equal to?
NDA Paper 1 2014

(A)

`log_e 5`

(B)

`log_5 e`

(C)

`5`

(D)

`1`

Solution:

`lim_(x -> 0) (5^x -1)/x`

`= log_e 5 quad (∵lim_(x -> 0) ( a^x - 1)/x = log_e a)`

Correct Answer is `=>` (A) `log_e 5`

Q 1732880732

If `f(9) = 9` and ` f' (9) = 4`, then what is `lim_(x -> 9) (sqrt (f(x)) - 3)/(sqrt(x) - 3)`
equal to?
NDA Paper 1 2014

(A)

`36`

(B)

`9`

(C)

`4`

(D)

None of these

Solution:

Given that,

`f(9) = 9` and `f' (9) = 4`

`lim_(x -> 9) (sqrt (f(x)) - 3)/(sqrt(x) - 3)`

` [ ∵ f(9) = 9 => f(x) = x. i.e., x = 9]`

Using `L`' Hospital rule;

`lim_(x -> 9) ( 1/(2sqrt(f(x))) . f'(x))/(1/(2sqrt(x)) .1)`

` lim_(x -> 9) (f' (x) xx sqrt(x))/( sqrt(f(x))`

` = (f'(9) xx sqrt(9))/( sqrt(f(9))) = (4 xx 3)/3 = 4`

Correct Answer is `=>` (C) `4`

Q 1773512446

What is ` lim_(x -> 0) (( 1 + x)^n -1)/x ` equal to?
NDA Paper 1 2014

(A)

`0`

(B)

`1`

(C)

`n`

(D)

`n - 1`

Solution:

` lim_(x -> 0) {(( 1 + x)^n -1)/x }`

` = lim_(x -> 0) (( 1 + nx + (n (n - 1))/(2!) x^2 + .....) - 1)/x `

(by binomial expansion in any index)

` = lim_(x -> 0) ( nx + (n(n -1))/(2!) x^2 + .......)/x`

` = lim_(x -> 0) n + (n(n -1))/(2!) x + ......`

` = n + 0+ 0+ ..... = n`

Correct Answer is `=>` (C) `n`

Q 1773612546

What is `lim_(x -> 0) x/sqrt(1 - cos x)` equal to?
NDA Paper 1 2014

(A)

` sqrt(2)`

(B)

`- sqrt(2)`

(C)

` 1/sqrt(2)`

(D)

Does not exist

Solution:

`lim_(x -> 0) x/sqrt(1 - cos x)`

` = lim_(x -> 0) x/sqrt(1 - (1 - 2 sin^2 (x/2))`

` = lim_(x -> 0) x/sqrt(2 sin^2 (x/2))`

` = 1/sqrt(2) lim_(x -> 0) x/(| sin (x/2)|)`

`LHL = f(0 - 0) = lim_(h -> 0) f(0 - h)`

` = 1/sqrt(2) lim_(h -> 0) (-h)/(|sin ((-h)/2)|)`

` = - 1/sqrt(2) lim_(h -> 0) (h)/(|-sin ((h)/2)|) = - 1/sqrt(2) lim_(h -> 0) (2.(h/2))/(sin (h/2))`

` = - 1/sqrt(2) xx 2 xx 1 quad ( ∵ lim_(theta -> 0) theta/(sin theta) = 1)`

`RHL= f (0 +0) = lim_(h -> 0) f(0+h)`

` = 1/sqrt(2) lim_(h -> 0) (h)/(|sin ((h)/2)|)`

` = 1/sqrt(2) lim_(h -> 0) (2(h/2))/(sin(h/2))`

` = 1/sqrt(2) xx 2 xx 1 quad ( ∵ lim_(theta -> 0) theta/(sin theta) = 1)`

` = sqrt(2)`

` ∵ LHL != RHL`

So, limit does not exist.

Correct Answer is `=>` (D) Does not exist

Q 1763456345

Consider the function

` f(x) = ( 1 - sin x)/(pi - 2x)^(2')`

where `x != pi/2 ` and ` f (pi/2) = lamda`

What is ` lim_(x -> pi/2) f(x)` equal to?
NDA Paper 1 2014

(A)

`1`

(B)

`1/2`

(C)

`1/4`

(D)

`1/8`

Solution:

Given function, ` f(x) = ( 1 - sin x)/(pi - 2x)^(2)`

where `x != pi/2 ` and ` f (pi/2) = lamda`

` lim_(x -> pi/2) f(x) = lim_(x -> pi/2) ( 1 - sin x)/(pi - 2x)^(2)`

Use 'L' Hospital rule,

` = lim_(x -> pi/2) (-cos x)/(2(pi-2x)(-2))`

` = lim_(x -> pi/2) (cos x)/(4 (pi - 2x))`

Again, use 'L' Hospital rule,

` = lim_(x -> pi/2) (- sin x) /(4 (-2)) = lim_(x -> pi/2) (sin x)/8`

` 1/8 . sin (pi/2) = 1/8 xx 1 = 1/8`

Correct Answer is `=>` (D) `1/8`

Q 2420112011

What is `lim_(x->2) (2-x)/(x^3 - 8)` equal to ?
NDA Paper 1 2013

(A)

`1/8`

(B)

`-1/8`

(C)

`1/12`

(D)

`-1/12`

Solution:

`lim_(x->2) (2-x)/(x^3 -8) = lim_(x->2) (-(x-2))/((x-2)(x^2 +4 +2x))`

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

`= (-1)/(4+4+4) = -1/12`

Correct Answer is `=>` (D) `-1/12`

Q 2480412317

What is `lim_(x->0) (1-cos x)/x` equal to?
NDA Paper 1 2013

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

`lim_(x->0) (1- cos x)/x` (form `0/0`)

Use L' Hospital rule, `lim_(x->0) (sin x)/1 = sin 0 =0`

Correct Answer is `=>` (A) `0`

Q 2450612514

What is `lim_(x->0) (cos x)/(pi-x)` equal to?
NDA Paper 1 2013

(A)

`0`

(B)

`pi`

(C)

`1/pi`

(D)

`1`

Solution:

`lim_(x->0) (cos x)/(pi -x) = (cos 0)/(pi - 0) = 1/pi`

Correct Answer is `=>` (C) `1/pi`

Q 2400612518

What is `lim_(x->0) (sin 2x +4x)/(2x + sin 4x)` equal to?
NDA Paper 1 2013

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`2`

Solution:

`lim_(x->0) (sin 2 x + 4x)/(2x + sin 4x)` (form `0/0`)

Use L' Hospital rule,

`= lim_(x->0) (2 cos 2x +4)/(2+ 4 cos 4x) = (2 cos 0 +4)/(2+4 cos 0)`

`= (2 xx 1 +4)/(2+4 xx 1) = (2+4)/(2+4) = 6/6 =1`

Correct Answer is `=>` (C) `1`

Q 2460723615

Consider the following statements

I. `lim_(x->0) sin (1/x)` does not exist.

II. `lim_(x->0) x sin ( 1/x)` exist.

Which of the above statements is/are correct?
NDA Paper 1 2013

(A)

Only I

(B)

Only II

(C)

Both 1 and II

(D)

Neither I nor II

Solution:

I. `lim_(x->0) sin (1/x)`

`LHL = f(0-0) = lim_(h->0) f(0-h)= lim_(h->0) sin (1/(-h))`

`= lim_(h->0) - sin (1/h) = - sin(oo)`

`=-(A` definite number persist between `-1 ` to `+ 1)`

`RHL = f (0+0) = lim_(h->) (0+h) = lim_(h->0) sin (1/h)`

`= sin (oo)`

`= (A` definite number persist between `-1` to `+ 1)`

`:. LHL ne RHL`

Hence, `lim_(x->0) sin (1/x)` does not exist.

II. `lim_(x->0) x sin (1/x)`

`LHL= f(0-0) = lim_(h->0)f(0-h) = lim_(h->0) (-h) sin (-1/h)`

`= lim_(h->0) h sin 1/h = 0 xx (A ` finite value persist between `-1` to `+ 1)`

`=0`

`RHL = f (0+0) = lim_(h->0) f(0+h)`

`= lim_(h->0) (h) sin (1/h)`

`= 0 xx sin (oo)`

`= 0 xx (A` finite value persist between `-1` to `+ 1)`

`=0`

`:. LHL = RHL`

Hence, `lim_(h->0) x sin (1/x)` exists.

Correct Answer is `=>` (C) Both 1 and II

Q 2410723619

Which is `lim_(x->0) (sin x - tan x)/x` equal to ?
NDA Paper 1 2013

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

`1/2`

Solution:

`lim_(x->0) (sin x - tan x)/x`

which is of the form `(0/0)`

Now, applying `L'` Hospital rule

`= lim_(x->0) (cos x - sec^2 x)/1 = (cos 0 - sec^2 0)/1 = (1-1)/1`

`=0/1 = 0`

`text (Alternate Method)`

`lim_(x->0) (sin x - tan x)/x= lim_(x->0) ( (x- x^3/(3!) + x^5/(5!) - ............ )- (x + x^3/3 +2/15 x^5 + .......) )/(x)`

`= lim_(x->0) (- (1/6 +1/3) x^3 + (1/120 -2/15) x^5 + ..........)/x`

`= lim_(x->0) -1/2 x^2 + ( (1-16)/120 ) x^4 + ..........`

` = -1/2 * 0 -1/8 *0 = 0`

Correct Answer is `=>` (A) `0`

Q 2430123912

What is the value of `lim_(x->0) (1- sqrt (1+x) )/x` ?
NDA Paper 1 2013

(A)

`1/2`

(B)

`-1/2`

(C)

`1`

(D)

`-1`

Solution:

`lim_(x->0) (1- sqrt (1+x) )/x = lim_(x->0) (1- sqrt (1+x) )/x * (1+ sqrt (1+x) )/(1+ sqrt (1+x) )` (rationalisation)

`= lim_(x->0) (1- [sqrt (1+x) ]^2)/(x (1+ sqrt (1+x) ) ) = lim_(x->0) (1- (1+x) )/(1+ sqrt (1+x) )`

`= lim_(x->0) (-x)/(x (1+ sqrt (1+x) ) ) = lim_(x->0) (-1)/(1 + sqrt (1+x) )`

`= (-1)/(1+ sqrt (1+0) ) = (-1)/(1+1) = -1/2`

Correct Answer is `=>` (B) `-1/2`

Q 2420234111

What is the value of `lim_(x->0) (sqrt (1+x) -1 )/x `?
NDA Paper 1 2012

(A)

`0`

(B)

`1/2`

(C)

`1`

(D)

`-1/2`

Solution:

Given, `lim_(x->0) (sqrt (1+x) -1 )/x`

`= lim_(x->0) (sqrt (1+x) -1 )/x * (sqrt (1+x) +1 )/(sqrt (1+x ) +1)` (rationalisation)

`= lim_(x->0) (1+x -1)/(x sqrt ( 1+x) +1) = lim_(x->0) x/(x (sqrt (1+x) +1 ) )`

`= lim_(x->0) 1/(1+ sqrt (1+x) ) = 1/(1+ sqrt (1+0) ) =1/2`

Correct Answer is `=>` (B) `1/2`

Q 2400334218

What is the value of `lim_(x->0) (2 (1- cos x) )/x^2` ?
NDA Paper 1 2012

(A)

`0`

(B)

`1/2`

(C)

`1/4`

(D)

`1`

Solution:

Given, `lim_(x->0) (2 (1- cos x) )/x^2 = lim_(x->0) (2 (1-1 +2 sin^2 x/2) )/x^2`

`[ :. cos x = 1-2 sin^2 x/2 ]`

`= lim_(x->0) (4 sin^2 x/2)/x^2 = lim_(x->0) ((sin x/2)/(x/2))^2 = (1)^2 =1` `( :. lim_(theta->0 ) (sin theta)/theta =1 )`

Correct Answer is `=>` (D) `1`

Q 2400434318

Consider the following statements

I. `lim_(x->0) 1/x ` exist .

II. `lim_(x->0) e^(1/x)` does not exist.

Which of the above statements is/are correct?
NDA Paper 1 2012

(A)

Only I

(B)

Only II

(C)

Both I and II

(D)

Neither I nor II

Solution:

I. Given, `lim_(x->0) = 1/x`

`LHL = f(0-0) = lim_(x->0) f (0-h) = lim_(x->0) - 1/h = -oo`

`RHL = f(0+0) = lim_(x->0) f(0+h) = lim_(x->0) + 1/h = +oo`

`:. LHL ne RHL`

Hence, `lim_(x->0) 1/x` does not exist.

II. Given, `lim_(x->0) e^(1/x)`

`LHL = f(0-0) lim_(h->0) f(0-h) = lim_(x->0) e^(-1/h)`

`= e^(-oo) = 0`

`RHL = f(0+0) = lim_(h->0) f(0+h) = lim_(h->0) e^(1/h) = e^(oo) = oo`

`:. LHL ne RHL`

So, `lim_(x->0) e^(1/x)` does not exist.

Hence, only Statement II is true.

Correct Answer is `=>` (B) Only II

Q 2440134913

What is the value of `lim_(x->2) (x-2)/(x^2 - 4)`?
NDA Paper 1 2012

(A)

`0`

(B)

`1/4`

(C)

`1/2`

(D)

`1`

Solution:

`lim_(x->2) (x-2)/(x^2 -4) = lim_(x->2) (x-2)/((x+2)(x-2))`

`= lim_(x->2) 1/(x+2) = 1/(2+2) = 1/4`

Correct Answer is `=>` (B) `1/4`

Q 2440345213

What is the value of `lim_(x->0) x^2 sin (1/x)` ?
NDA Paper 1 2012

(A)

`0`

(B)

`1`

(C)

`1/2`

(D)

Does not exist

Solution:

`lim_(x->0) x^2 sin (1/x)`

Put `x = 1y`,

`= lim_(y->oo) 1/y^2 * sin (y)`

`= lim_(y-> oo) (sin y)/y^2` `( :. -1 le sin theta le 1, AA theta in R )`

`= (sin oo)/(oo) = 0 xx (A` finite value persist between `-1` to `+ 1)`

`=0`

Correct Answer is `=>` (A) `0`

Q 2480445317

What is `lim_(x-> oo) (sqrt (a^2 x^2 +ax +1) - sqrt (a^2 x^2 +1) )` equal to?
NDA Paper 1 2011

(A)

`1/2`

(B)

`1`

(C)

`2`

(D)

`0`

Solution:

`lim_(x-> oo) ( sqrt (a^2 x^2 +ax +1) - sqrt (a^2 x^2 +1))`

After rationalisation,

`= lim_(x-> oo) ( (a^2x^2 +ax +1 - a^2x^2 -1)/( sqrt (a^2 x^2 +ax +1) + sqrt (a^2 x^2 +1)))`

`= lim_(x->oo) a/(sqrt (a^2 + a/x + 1/x^2) + sqrt (a^2 + 1/x^2) )`

`= (+a)/(sqrt (a^2) + sqrt (a^2) ) = a/(2a) = 1/2`

Correct Answer is `=>` (A) `1/2`

Q 2480145917

What is the value of `lim_(x->0) (cos (ax) -cos (bx) )/x^2` ?
Paper 1 2010

(A)

`a- b`

(B)

`a+b`

(C)

`(b^2 - a^2)/2`

(D)

`(b^2 + a^2)/2`

Solution:

`lim_(x->0) (cos ax - cos bx)/x^2` (`0/0` form)

Using L' Hospital rule,

`= lim_(x->0) (-a sin ax + b sin bx)/(2x)`

Using L' Hospital rule,

`= lim_(x->0) (-a^2 cos ax + b^2 cos bx)/2` (`0/0` form)

`= (-a^2 cos 0 + b^2 cos 0)/2 = (-a^2 (1) + b^2 (1) )/2`

`= (b^2 -a^2)/2`

Correct Answer is `=>` (C) `(b^2 - a^2)/2`

Q 2400145918

What is the value of `lim_(x->0) (cos (ax) -cos (bx) )/x^2` ?
NDA Paper 1 2010

(A)

`a- b`

(B)

`a+b`

(C)

`(b^2 - a^2)/2`

(D)

`(b^2 + a^2)/2`

Correct Answer is `=>` (C) `(b^2 - a^2)/2`

Q 2410156010

`lim_(x->0) (a^x - b^x)/x` is equal to?
NDA Paper 1 2010

(A)

`log (ab)`

(B)

`(log a)/(log b)`

(C)

`log (a/b)`

(D)

`log (b/a)`

Solution:

`lim_(x-> 0) (a^x - b^x)/x` (`0/0` form)

`= lim_(x->0) (a^x log a - b^x log b)/1` (by L' Hospital rule )

`= (a^0 log a - b^0 log b)/1 = ((1) log a - (1) log b)/1`

`= log a - log b = log a/(logb)`

Correct Answer is `=>` (C) `log (a/b)`

Q 2440256113

What is the value of `lim_(x-> oo) ((x+6)/(x+1))^(x+4)` ?
NDA Paper 1 2010

(A)

`e`

(B)

`e^2`

(C)

`e^4`

(D)

`e^5`

Solution:

`lim_(x->oo) ((x+6)/(x+1))^(x+4)`

`= lim_(x->oo) (1+5/(x+1))^( (x+4)/5 * 5/(x+1) * (x+1) )`

`= lim_(x->oo) [ (1+5/(x+1))^((x+1)/5)]^(5* (x+4)/(x+1))` [ `(1)^(oo)` form ]

`= e^(5 lim_(x->oo) (1+4/x)/ (1+1/x))` `[ :. lim_(x->oo) (1+1/x)^x = e ]`

`= e^5`

Correct Answer is `=>` (D) `e^5`

Q 2470256116

What is the value of `lim_(x->1) (x-1)^2/(|x-1|)`?
NDA Paper 1 2010

(A)

`0`

(B)

`1`

(C)

`-1`

(D)

Does not exist

Solution:

Let `f(x) = (x-1)^2/(| x-1 |)`

Redefined the function,

`=> f(x) = { tt ( ( (x-1), x ge 1 ), (-(x-1) , x < 1) )`

Now, `LHL = lim_(h->0) f (1-h)`

`= lim_(h->0) [- (1-h -1) ] = lim_(h->0) h =0`

and `RHL = lim_(h->0) f(1+h)`

`= lim_(h->0) (1+h -1) = lim_(h-> 0) h =0`

`:. LHL = RHL`

`=> lim_(x->1) (x-1)^2/(|x-1 |)= 0`

Correct Answer is `=>` (A) `0`

Q 2420356211

What is `lim_(x->oo) (x/(3+x))^(3x)` equal to?
NDA Paper 1 2009

(A)

`e`

(B)

`e^3`

(C)

`e^(-9)`

(D)

`e^9`

Solution:

`lim_(x->oo) (x/(3+x))^(3x) = lim_(x->oo) ((3+x)/x)^(-3x)`

`= lim_(x->oo) (1+3/x)^(-3x) = lim_(x->oo) (1+3/x)^(-3x)`

`= lim_(x->oo) (1+3/x)^(x/3 (-9) )` [`(1)^(oo)` form]

`[ :. lim_(x->oo) (1+1/x)^x = e ]`

`= e^(-9)`

Correct Answer is `=>` (C) `e^(-9)`

Q 2480456317

What is `lim_(x->0) (sin^2 ax)/(bx)` ( where, `a` and `b` are constants ) equal to?
NDA Paper 1 2009

(A)

`0`

(B)

`a`

(C)

`a^2/b`

(D)

Does not exist

Solution:

`lim_(x->0) (sin^2 ax)/(bx) * a^2/a^2 * x/x`

`= lim_(x-> 0) ((sin ax)/(ax))^2 * a^2/b x =1 * a^2/b * 0 = 0`

`( :. lim_(theta->0) (sin theta)/(theta) =1)`

Correct Answer is `=>` (A) `0`

Q 2460556415

`lim_(x->0) e^(-1/x)` is equal to
NDA Paper 1 2008

(A)

`0`

(B)

`oo`

(C)

`e`

(D)

Does not exist

Solution:

`LHL = lim_(h->0) e^(-1/(0-h))`

`= lim_(h)->0 e^(1/h) = e^(oo) = oo`

`RHL = lim_(h->0) e^(-1/(0+h))`

`= lim_(h->0 ) e^(-1/h) = e^(-oo) = 0`

`:. LHL ne RHL`

So , `lim_(h->0) e^(-1/x)` does not exist.

Correct Answer is `=>` (D) Does not exist

Q 2470656516

If `f(x) = 1/(1- | 1-x |)` then what is `lim_(x->1) f(x)` equal to?
NDA Paper 1 2008

(A)

`0`

(B)

`oo`

(C)

`1`

(D)

`-1`

Solution:

`LHL = lim_(h->0) 1/(1- | 1- (1-h) |) = lim_(h->0) 1/(1- |h|)`

`= lim_(h->0) 1/(1-h) =1`

`RHL = lim_(h->0) 1/(1- | 1- (1+h) |)`

`= lim_(h->0) 1/(1- | -h |) = lim_(h->0) 1/(1-h) =1`

`:. lim_(x->1) f(x) = 1`

Correct Answer is `=>` (C) `1`

Q 2420756611

What is the value of `lim_(x->alpha) ( sqrt(alpha +2x) - sqrt (3x))/( sqrt (3 alpha +x) - 2 sqrt (x) )` ?
NDA Paper 1 2008

(A)

`2/sqrt(3)`

(B)

`1/(3 sqrt (3))`

(C)

`2/(3 sqrt(3))`

(D)

`1/sqrt(3)`

Solution:

`lim_(x-> alpha) ( sqrt (alpha +2x) - sqrt (3x))/(sqrt (3 alpha +x) -2 sqrt (x))`

After rationalisation,

`= lim_(x->alpha) ( (sqrt (alpha +2x) )^2 - (sqrt (3x))^2 )/( sqrt (alpha +2x) + sqrt (3x)) * (sqrt (3 alpha +x) +2 sqrt (x))/( (sqrt (3 alpha +x) )^2 - (2 sqrt(x))^2)`

`= lim_(x->alpha) (alpha +2x -3x)/(sqrt (alpha +2x + sqrt (3x)) ) * (sqrt (3 alpha +2x) + 2 sqrt(x) )/(3 alpha +x -4x)`

`= lim_(x->alpha) (sqrt (3 alpha +x) + 2 sqrt (x) )/(sqrt (alpha +2x ) + sqrt (3x) ) * (alpha -x)/(3 (alpha -x) )`

`= 1/3 lim_(x->alpha) (sqrt (3 alpha +x) + 2 sqrt (x) )/(sqrt (alpha + 2x) + sqrt (3x) )`

`= 1/3 * ( sqrt (4 alpha) + 2 sqrt (alpha) )/(sqrt (3 alpha) + sqrt(3 alpha) ) = 1/3 * (4 sqrt (alpha) )/(2 sqrt (3) sqrt (alpha)) = 2/(3 sqrt (3)) .`

Correct Answer is `=>` (C) `2/(3 sqrt(3))`

Q 2440756613

What is the value of `lim_(x->oo) (sin x)/x ?`
NDA Paper 1 2008

(A)

`1`

(B)

`0`

(C)

`oo`

(D)

`-1`

Solution:

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

Put `x =1/n` and as `x-> oo , h -> 0`

`:. lim_(x->oo) (sin x)/x = lim_(h->0) h.sin 1/h = 0 xx ` (finite value)

`=0 ` (Since, value of `sin 1/h` lies between `-1` and `1` )

Correct Answer is `=>` (B) `0`

Q 2400278118

What is the value of `lim_(x->oo) ((x-2)/(x+2))^(x+2)`?
NDA Paper 1 2007

(A)

`0`

(B)

`e^4`

(C)

`e^(-2)`

(D)

`e^(-4)`

Solution:

`lim_(x->oo) ((x-2)/(x+2))^(x+2)`

`= [lim_(x->oo ) [ 1+ ((-4)/(x+2)) ]^((-(x+2))/(4))]^(-4)` [ `(1)^(oo) ` form ]

` [ :. lim_(x->oo ) (1+1/x)^x = e ]`

`= e^(-4)`

Correct Answer is `=>` (D) `e^(-4)`