Question

Evaluate :


$\displaystyle\lim_{x \to 0}\: {(1+x )}^{\frac{1}{x}}$

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Answer 1

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Answer 2

ANSWER

$e^{1}$or e

Step by step Explanation

We have ->

$\displaystyle\lim_{x \rightarrow 0}\: {(1+x )}^{\frac{1}{x}}$

$Now \:this \:is \:of \:the\: form\: (\rightarrow )^{\rightarrow \infty}$

$[Since \displaystyle\lim_{x \rightarrow a}{f(x)}^{g(x)} = e^{\displaystyle\lim_{x\rightarrow 0} }\:(f(x) - 1)g(x)]$

Now, we have →

$e^{\displaystyle\lim_{x \rightarrow 0} }(1 + x - 1)\dfrac{1}{x}$

$\implies e^{\displaystyle\lim_{x \rightarrow 0} }(x)\dfrac{1}{x}$

Now we know that

$x \times \dfrac{1}{x}$ = 1

On putting this in the expression we get→

$\implies e^{\displaystyle\lim_{x \rightarrow 0} (1) }$

$\implies e^{1}$

This is equal to e.

Hence the value if the expression is $e^{1}$ or e.