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Find the value of $\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$

We can write this limit function as :

$$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{\large{x^2}}}$$

Please guide further how to proceed in such limit..

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4 Answers

up vote 5 down vote accepted

Write it as $$\dfrac{(1+5x^2)^{1/x^2}}{(1+3x^2)^{1/x^2}}$$ and recall that $$\lim_{y \to 0} (1+ay)^{1/y} = e^a$$ to conclude what you want.

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Let $n = \displaystyle\frac{1}{x^{2}}$:

$\displaystyle \lim_{x \rightarrow 0}\left(\frac{1+5x^2}{1+3x^2}\right)^{\frac{1}{\large {x^2}}}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(\frac{n + 5}{n + 3}\right)^{n}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(1 + \frac{2}{n + 3}\right)^{n}$ = $\displaystyle\lim_{n \rightarrow \infty}\left(1 + \frac{2}{n}\right)^{n}$ = $e^{2}$

by definition of $e$.

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$$\lim_{x \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}$$

$$=\lim_{x \rightarrow 0}\left(\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge{\frac2{1+3‌​x^2}}}$$

$$=\left(\lim_{\frac{2x^2}{1+3x^2} \rightarrow 0}\left(1+ \frac{2x^2}{1+3x^2}\right)^{\huge{\frac{1+3x^2}{2x^2}}}\right)^{\huge\lim_{x \rightarrow 0}{\frac2{1+3‌​x^2}}}=e^2$$

as $x\to0\implies \frac{2x^2}{1+3x^2}\to0$

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@GitGud, thanks for your help. –  lab bhattacharjee Apr 21 '13 at 17:52
No problem. I suggest that you use \large for the limits instead of \huge. Also try \lim \limits_{x\to 0} instead of \lim _{x\to 0}. –  Git Gud Apr 21 '13 at 17:54
@GitGud, btw, do you know how to wrap text in the cell of a table using latex? –  lab bhattacharjee Apr 21 '13 at 17:56
I don't understand what you want. –  Git Gud Apr 21 '13 at 17:57
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Based on the fact that

$$\forall\,a\in\Bbb R\;,\;\;\lim_{x\to x_0}\left(1+\frac{a}{f(x)}\right)^{f(x)}=e^a\;,\;\;\text{whenever $\,f\,$ is a function s.t.}\;\;f(x)\xrightarrow[x\to x_0]{}\infty$$

we get:

$$\left(1+ \frac{2x^2}{1+3x^2}\right)^{\frac{1}{x^2}}=\left(1+ \frac{2}{3+\frac{1}{x^2}}\right)^{3+\frac{1}{x^2}}\left(1+ \frac{2}{3+\frac{1}{x^2}}\right)^{-3}\xrightarrow[x\to0]{}e^2\cdot 1=e^2$$

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