How do you solve this ? $lim_(x->0^+)(1+sin(4x))^cot(x)$

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Answer 1 · 2017-08-08T22:03:21

$lim_{x\to0^+} (1+sin(4x))^(cot(x)) = e^4$ .

$y=(1+sin(4x))^(cot(x))$
$ln(y)=cot(x)ln(1+sin(4x)$ ,
$ln(y)=ln(1+sin(4x))/(tan(x))$ .

Then $lim_{x\to0^+} ln(y)$ is in the indeterminate form $0/0$ .

By L'Hopitals rule, if $f(a)=g(a)=0$ then $lim_{x\toa} (f(a))/(g(a)) = lim_{x\toa} (f'(a))/(g'(a))$ .

Then,

$lim_{x\to0^+} ln(y) = lim_{x\to0^+} ((4cos(4x))/(1+sin(4x)))/(sec^2(x))$ ,
$lim_{x\to0^+} ln(y) = 4$ .

Then taking an inverse logarithm,

$lim_{x\to0^+} y = e^4$ .

How do you solve this ? lim_(x->0^+)(1+sin(4x))^cot(x)lim_(x->0^+)(1+sin(4x))^cot(x)