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Phrases Limit PYQ


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lim, 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}}


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Let f(x)=\frac{x^2-1}{|x|-1}. Then the value of lim_{x\to-1} f(x) is





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 is equal to 





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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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Which of the following is NOT true?





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





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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}


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





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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


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The value of \lim_{x\to a} \frac{\sqrt{a+2x}-\sqrt{3x}}{\sqrt{3a+x}-2\sqrt{x}}





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What is the value of \lim _{{x}\rightarrow\infty}-(x+1)\Bigg{(}{e}^{\frac{1}{x+1}}-1\Bigg{)}?





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Function is the form of  therefore using by L'Hospital rule
Again apply L'Hospital Rule,
Putting x = 0, we get 
 


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Find 





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If 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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