Phrases Limit Previous Year Question Paper and Solution with PDF

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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Qus : 8Phrases PYQ

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

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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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Phrases Previous Year PYQPhrases NIMCET 2017 PYQ

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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 : 11Phrases PYQ

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Phrases Previous Year PYQPhrases NIMCET 2017 PYQ

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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

Qus : 12Phrases PYQ

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Qus : 13Phrases PYQ

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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