Mathematics Must Do Problems of Limits for NDA

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Must Do Problems of Limits for NDA

Must do problem for NDA

Q 1851112924

If ` f(x) = ( ((sin[x])/([x]), [x] != 0), ( 0, [x] = 0 ))` where `[.]` denotes the greatest integer

function, then `lim_( x -> 0) f(x)` is equal to
NCERT Exemplar

(A)

`1`

(B)

`0`

(C)

`-1`

(D)

Does not exist

Solution:

Given, ` f(x) = { ((sin[x])/([x]), [x] != 0), ( 0, [x] = 0 ))`

` :. LHL = lim_( x -> 0^-) f(x)`

`= lim_( x -> 0^-) ( sin [x])/([x]) = lim_( h -> 0) ( sin [ 0 - h]) /( [ 0 - h]) `

` = sin1`

` RHL = lim_( x -> 0^+) f(x) = lim_( x -> 0^+) ( sin [x])/([x]) `

` = lim_( x -> 0^+) ( sin [ 0 + h]) /( [ 0 + h]) = lim_( h -> 0) ( sin [ h])/([ h]) = 1`

` ∵ LHL!= RHL`

So, limit does not exist.

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

Q 1326112971

`lim_{x to 0}frac{ x^n-[x]}{[x]}n in N` [`[x]` denotes greatest integer less than or equal to `x`]

(A)

has value `-1`

(B)

has value `0`

(C)

has value `1`

(D)

does not exist

Solution:

`LHL lim_{x to 0^-}[x] =-1`

`RHL lim_{x to 0^+}[x] =0`

`LHL ne RHL`

Since `lim_{x to 0}[x]` does not exist,hence required limit does not exist.

Hence `4` is the correct answer.

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

Q 1811423320

If `f(x) = {((x^2 - 1, 0 < x < 2),( 2x + 3, 2 <= x < 3 ))` then the quadratic equation whose roots

are `lim_(x -> 2^-) f(x)` and `lim_( x -> 2^+) f(x)` is
NCERT Exemplar

(A)

`x^2 - 6x + 9 = 0`

(B)

`x^2 - 7x + 8 = 0`

(C)

`x^2 - 14x + 49 = 0`

(D)

`x^2 - 10 x + 21 = 0`

Solution:

Given, `f(x) = {((x^2 - 1, 0 < x < 2),( 2x + 3, 2 <= x < 3 ))`

`:. lim_(x -> 2^-) f(x) = lim_(x -> 2^-) (x^2- 1)`

` = lim_(h -> 0) [(2- h)^2- 1] = lim_(h -> 0) (4+ h^2 - 4h - 1)`

` = lim_(h -> 0) (h^2 - 4h + 3) = 3`

and ` = lim_(x -> 2^+) f(x) = lim_(x -> 2^+) (2x + 3)`

`= lim_(h -> 0) [2(2 + h)+ 3] = lim_(h -> 0) (4+ 2h+ 3) = 7`

So, the quadratic equation whose roots are `3` and `7` is `x^2 - (3 + 7)x + 3 xx 7 = 0 ` i.e.,

` x^2 - 10x + 21 = 0` .

Correct Answer is `=>` (D) `x^2 - 10 x + 21 = 0`

Q 2814201159

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

(A)

(B)

(C)

(D)

does not exist

Solution:

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

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

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

= does not exist

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

Q 2853856744

`lim_(theta -> pi/4) (sqrt 2 - cos theta - sin theta )/(4 theta - pi )^2 ` is equal to

(A)

`1/16`

(B)

`1/(16 sqrt 2)`

(C)

`1/8`

(D)

`1/(8 sqrt 2)`

Solution:

`lim_(theta -> pi/4) (sqrt 2 - sqrt 2 cos (theta - pi/4) )/(16 (theta - pi/4)^2 )`

`= lim_(y->0) (sqrt 2)/16 * (1- cos y)/( y^2 )`

where, `y = theta - pi/4 -> 0 ` as `theta -> pi/4`

`=1/(8 sqrt 2) * lim_(y->0) (2 sin^2 (y/2) )/y^2`

`= 1/(8 sqrt 2) * 1/2 = 1/(16 sqrt 2) [ :. lim_( theta->0) sin theta = 0 ]`

Correct Answer is `=>` (B) `1/(16 sqrt 2)`

Q 2569778615

The value of ` lim_(n -> oo ) { 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/( (2n + 1)(2n + 3)) }` is
BCECE Mains 2015

(A)

`1`

(B)

`1//2`

(C)

`- 1//2`

(D)

None of the above

Solution:

We have,

` lim_(n -> oo ) { 1/(1.3) + 1/(3.5) + 1/(5.7) + ... + 1/( (2n + 1)(2n + 3)) }`

` = 1/2 lim_(n -> oo ) { ( 1 - 1/3) + ( 1/3 - 1/5) + ( 1/5 - 1/7) + ... + ( 1/(2n + 1) - 1/(2n + 3) ) }`

` = 1/2 lim_(n -> oo ) { 1 - 1/(2n + 3) } = 1/2 lim_(n -> oo ) ( 2n + 2)/(2n + 3) = 1/2`

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

Q 2259880714

`Lim_(x->1) (sqrt (1-cos 2 (x-1) ) )/ (x-1)`
BITSAT Mock

(A)

exists and it equals `sqrt(2)`

(B)

exists and it equals ` -sqrt(2)`

(C)

does not exist because `x - 1 -> 0`

(D)

does not exist because left hand limit is not equal to right hand limit.

Solution:

`Lim_(x->1) (sqrt (1-cos 2 (x-1)) )/(x-1)`

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

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

`= { tt ( (sqrt (2) \ \ \ \ \text(if) \ \ \ \ \y-> 0 +), (- sqrt(2) \ \ \ \ text(if) \ \ \ \ \y -> 0 -) )`

Hence the left hand and right hand limits are unequal and
therefore limit does not exist.

Correct Answer is `=>` (D) does not exist because left hand limit is not equal to right hand limit.

Q 1366112075

`lim_{x to 0}frac{sin x-x}{x^3}=`

(A)

`frac{1}{3}`

(B)

`frac{-1}{3}`

(C)

`frac{1}{6}`

(D)

`frac{-1}{6}`

Solution:

Substituting the `x` by `0`.

As we can see this is `0/0` indeterminate form, hence we use L'Hospital's rule

`lim_{x to 0} frac{\sin x-x}{x^3}`

`Rightarrow lim _{x to 0}frac{cos x-1}{3x^2}`

`Rightarrow lim_{x to 0}frac{-sin x}{6x}`

`Rightarrow -frac{1}{6}`

Hence `4` is the correct answer.

Correct Answer is `=>` (D) `frac{-1}{6}`

Q 2844112953

If `lim_(x->a) f(x) = 1` and

`lim_(x->a) g(x) = oo`, then `lim_(x->a) [f(x)]^(g(x)) = e^( lim_(x->a) [ f(x) -1] g(x) )` then

`lim_(x->0) (cos x )^(cot^2 x)` equals

(A)

(B)

`e^2`

(C)

`e^(-1/2)`

(D)

Solution:

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

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

`= e^(-1/2) [ :. lim_(x->0) (1- cos x)/(x^2) =1/2 ]`

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

Q 1356001874

If `f(5)=15,f'(5)=3` then `lim_{x to 5}frac{xf(5)-5f(x)}{x-5}` is

(A)

`15`

(B)

`0`

(C)

`3`

(D)

None of these

Solution:

Substituting the `x` by `5`.

As we can see this is `0/0` indeterminate form, hence we use L'Hospital's rule

`lim_{x to 5} frac{xf(5)-5f(x)}{x-5}`

`Rightarrow lim_{x to 5}frac{f(5)-5f'(x)}{1}`

Substituting the `x` by `5`.

`Rightarrow frac{15-5 times 3}{1}`

`Rightarrow 0`

Hence `2` is the correct answer.

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

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 1316123070

`lim_{x to infty}frac{x^n}{e^x}=0` for

(A)

`n=0` only

(B)

`n` is any whole number

(C)

`n=2` only

(D)

No value of `n`

Solution:

`lim_{x to infty}frac{x^n}{e^x}`

`(frac{infty}{infty})`

`L` Hospital rule,

`Rightarrow lim_{x to infty}frac{nx^{n-1}}{e^x}`

`Rightarrow lim_{x to infty}frac{n!}{e^{ x}}=0`

Where `n` is any whole number as `n!` is defined for the integers and `0`.

Hence `2` is the correct answer.

Correct Answer is `=>` (B) `n` is any whole number

Q 1316812770

`lim_{x to a}frac{log(x-a)}{log(e^x-e^a)}` is

(A)

`1`

(B)

`-1`

(C)

`0`

(D)

None of these

Solution:

Substituting the `x` by `a`.

As we can see this is `infty /infty ` indeterminate form, hence we use L'Hospital's rule

Using `L` Hospitals rule `(frac{infty}{infty})`

`Rightarrow lim_{x to a}frac{1}{x-a} / frac{e^x}{e^x-e^a}`

`Rightarrow lim_{x to a}frac{1}{x-a} . frac{e^x-e^a}{e^x}`

Still it remains indeterminate form, so differentiating again.

`Rightarrow lim_{x to a}frac{e^x}{e^x+(x-a)e^{ x}}`

Substituting `x=a` in the above expression,

`Rightarrow frac{e^a}{e^a}`

`Rightarrow 1`

Hence `1` is the correct answer.

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

Q 1672780636

`underset (x -> oo) lim (int_0^(2x) xe^(x^2) dx)/(e^((4x)^2))` equals
BITSAT 2015

(A)

`0`

(B)

`oo`

(C)

`2`

(D)

`1/2`

Solution:

`underset (x -> oo) lim (int_0^(2x) xe^(x^2) d x)/(e^((4x)^2)) = underset (x -> oo) lim (int_0^(2x) e^(x^2) d(x^2))/(2e^((4x)^2))`

` = underset (x -> oo) lim ([e^(x^2)]_0^(2x))/(2e^((4x)^2)) = underset (x -> oo) lim(e^((4x)^2)-1)/(2e^((4x)^2))`

`underset (x -> oo) lim 1/2 - 1/(2e^((4x)^2)) = 1/2`

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

Q 2459191014

`lim_(x->oo) ((x+a)/(x+b))^(x+b)` is
BCECE Stage 1 2012

(A)

`1`

(B)

`e^(b-a)`

(C)

`e^(a-b)`

(D)

`e^b`

Solution:

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

`= e^(lim_(x->oo )(1+(a-b)/(x+b) - 1 ).(x+b)`

` = e^(a-b)`

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

Q 2813280149

If `f(x) = (sin (e^(x-2) -1 ) )/( log (x-1) ) ` then `lim_(x->2) f(x)` is given by

(A)

-2

(B)

-1

(C)

(D)

Solution:

Let `e^( x -2) - 1 = theta`, then as

`x -> 2 , theta -> 0`

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

`:. lim_(x->2) f(x) = lim_(x->2) (e^(x-2) -1 )/( log (x-1) )` [0/0 form]

Since, log 1 is 0.
Apply L'Hospital's rule,

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

`=e^0 * (2-1) = 1 * 1 = 1`

Correct Answer is `=>` (D) 1

Q 2683223147

Which of the following is/are true

(This question may have multiple correct answers)

(A) If `Lim_(x->a) {f(x) + g(x) }` exist , then both `Lim_(x->a) f(x)` and `Lim_(x->a) g(x)` exist

(B) If `Lim_(x->a) f(x)` and `Lim_(x->a) g(x)` exist , the `Lim_(x->a) {f(x) + g(x) }` exist

(D) If `Lim_(x->a) {f(x) g(x) } ` exist , then both `Lim_(x->a) f(x)` and `Lim_(x->a) g(x)` exist

Solution:

By standard results

Correct Answer is `=>` (B)