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Limit and Continuity

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Evaluate the given limit :

$lim_{x\rightarrow0} \ {\Large \frac{(1+x)^{\frac{1}{x}}-e}{x}}$

options :

$ \\ a) \frac{e}{8}  \\ b) -\frac{e}{2} \\ c)- \frac{e}{4}  \\ d) 1$
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$\Rightarrow (1+x)^{\Large{\frac{1}{x}}}=e^{\Large{\frac{1}{x}}{ln(1+x)}}=e^{\Large\frac{1}{x}(x-x^{2}/2+x^{3}/3.....)}$

$\Rightarrow e^{\Large({1-\frac{x}{2}+\frac{x^2}{3}- .....})}$

$\therefore \lim_{x\rightarrow0} \{e^{\Large(1-x/2+x^{2}/3.....)}-e\}/x$ 

Will produce  $\frac{0}{0}$ form.

Now apply L'hospital rule

$\Rightarrow - \color{Red}{\Large \frac{e}{2}}$

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