The correct option is A 1 Given, x → 0lim (tanxx) Apply L Hospital's rule =x → 0lim ( ^2(x)1) =x → 0lim ( ^2(x)) = ^2(0)=1 ...

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L'Hospital Rule to Remove Indeterminate Form

x → 0lim tan ...

Question

$l i m x \to 0 \frac{tan x}{x}$

1

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e^{\frac{1}{2}}

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e^{\frac{1}{3}}

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e

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Solution

The correct option is A $1$
Given,

$l i m x \to 0 (\frac{t a n x}{x})$

Apply L-Hospital's rule

$= l i m x \to 0 (\frac{{sec}^{2} (x)}{1})$

$= l i m x \to 0 ({sec}^{2} (x))$

$= {sec}^{2} (0) = 1$

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