The correct option is C 1 Lt_x→ 0^+(sinx)^tan x log k =Lt_ x→ 0^+ tan x logsin x #160; k= e ^ Lt_ x→ 0^+ tan x logsin x #160; #160; ...

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Substitution Method to Remove Indeterminate Form

Ltx→ 0+sinxta...

Question

$L t_{x \to 0^{+}} (s i n x)^{tan x} =$

e

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e^{2}

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- 1

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Solution

The correct option is C 1
$L t_{x \to 0^{+}} (s i n x)^{tan x}$
$log k = L t_{x \to 0^{+}} tan x log sin x$
$k = e^{L t_{x \to 0^{+}} tan x log sin x}$
$k = e^{L t_{x \to 0^{+}} \frac{log sin x}{cot x}}$
It is of the form $\frac{\infty}{\infty}$ , so applying L-Hospital's rule
$k = e^{L t_{x \to 0^{+}} \frac{cot x}{- {csc}^{2} x}}$
$k = e^{0} = 1$

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