Phrases Limit Previous Year Question Paper and Solution with PDF

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Phrases Previous Year Questions (PYQs)

Phrases Limit PYQ

Qus : 1

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Solution

Qus : 2

$\lim _{{x}\rightarrow1}\frac{{x}^4-1}{x-1}=\lim _{{x}\rightarrow k}\frac{{x}^3-{k}^2}{{x}^2-{k}^2}=$ , then find k

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Solution

Quick DL Method Solution

Given:

$\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2}$

LHS using derivative:

$\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \left.\frac{d}{dx}(x^4)\right|_{x=1} = 4x^3|_{x=1} = 4$

RHS using DL logic:

$\lim_{x \to k} \frac{x^3 - k^2}{x^2 - k^2} \approx \frac{3k^2(x - k)}{2k(x - k)} = \frac{3k}{2}$

Equating both sides:

$\frac{3k}{2} = 4 \Rightarrow k = \frac{8}{3}$

$\boxed{k = \frac{8}{3}}$

Qus : 3

Let

$f(x)=\frac{x^2-1}{|x|-1}$ . Then the value of

$lim_{x\to-1} f(x)$ is

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Solution

Qus : 4

is equal to

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Solution

Qus : 5

The value of the limit

$\lim _{{x}\rightarrow0}\Bigg{(}\frac{{1}^x+{2}^x+{3}^x+{4}^x}{4}{\Bigg{)}}^{1/x}$ is

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Solution

Qus : 6

Which of the following is NOT true?

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Solution

Qus : 7

For

$a\in R$ (the set of al real numbers),

$a \ne 1$ ,

$\lim _{{n}\rightarrow\infty}\frac{({1}^a+{2}^a+{\ldots+{n}^a})}{{(n+1)}^{a-1}\lbrack(na+1)(na+b)\ldots(na+n)\rbrack}=\frac{1}{60}$ . Then one of the value of

$a$ is

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Solution

Qus : 8

The value of

${{Lt}}_{x\rightarrow0}\frac{{e}^x-{e}^{-x}-2x}{1-\cos x}$ is equal to

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Solution

Evaluate:

$\lim_{x \to 0} \frac{e^x - e^{-x} - 2x}{1 - \cos x}$

Step 1: Apply L'Hôpital's Rule (since it's 0/0):

First derivative:

$\frac{e^x + e^{-x} - 2}{\sin x}$

Still 0/0 → Apply L'Hôpital's Rule again:

$\frac{e^x - e^{-x}}{\cos x}$

Now,

$\lim_{x \to 0} \frac{1 - 1}{1} = 0$

Final Answer:

$\boxed{0}$

Qus : 9

$\lim_{x\to \infty} (\frac{x+7}{x+2})^{x+5}$ equal to

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Solution

Qus : 10

Let f(x) be a polynomial of degree four, having extreme value at x = 1 and x = 2. If

$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$ , then f(2) is

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Solution

Given it has extremum values at x=1 and x=2
⇒f′(1)=0 and f′(2)=0
Given f(x) is a fourth degree polynomial
Let

$f(x)=a{x}^4+b{x}^3+c{x}^2+dx+e$

Given

$\lim _{{x}\rightarrow0}[1+\frac{f(x)}{{x}^2}]=3$

$\lim _{{x}\rightarrow0}\lbrack1+\frac{a{x}^4+b{x}^3+c{x}^2+\mathrm{d}x+e}{{x}^2}\rbrack=3$

$\lim _{{x}\rightarrow0}\lbrack1+a{x}^2+bx+c+\frac{d}{x}+\frac{e}{{x}^2}\rbrack=3$

For limit to have finite value, value of 'd' and 'e' must be 0

⇒d=0 & e=0

Substituting x=0 in limit

⇒ c+1=3

⇒ c=2

$f^{\prime}(x)=4a{x}^3+3b{x}^2+2cx+d$

$x=1$ and

$x=2$ are extreme values,

⇒

$f^{\prime}(1)=0$ and $f^{\prime}(2)=0

⇒

$4a+3b+4=0$ and

$32a+12b+8=0$

By solving these equations

we get,

$a=\frac{1}{2}$ and

$b=-2$

So,

$f(x)=\frac{x^{4}}{2}-2x^{3}+2x^{2}$

⇒

$f(x)=x^{2}(\frac{x^{2}}{2}-2x+2)$

⇒

$f(2)=2^{2}(2-4+2)$

⇒

$f(2)=0$

Qus : 11

The value of

$\lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}$

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Solution

Qus : 12

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Solution

Function is the form of

therefore using by L'Hospital rule

Again apply L'Hospital Rule,

Putting x = 0, we get

Qus : 13

Find

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Solution

Qus : 14

$f(x)=\lim _{{x}\rightarrow0}\, \frac{{6}^x-{3}^x-{2}^x+1}{\log _e9(1-\cos x)}$ is a real number then

$\lim _{{x}\rightarrow0}\, f(x)$

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Solution

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