lim_(x->0)(x/sinx)^(3/x^2)=  ?

Nov 12, 2016

Rewrite and use l'Hopital's Rule multiple times.

Explanation:

The initial form is indeterminate ${1}^{\infty}$.

A common way of dealing with this form is to rewrite as $e$ to a power, then find the limit of the exponent. By continuity of ${e}^{x}$, this lets us find the limit.

${\left(\frac{x}{\sin} x\right)}^{\frac{3}{x} ^ 2} = {e}^{\frac{3}{x} ^ 2 \ln \left(\frac{x}{\sin} x\right)}$

We want to find

${\lim}_{x \rightarrow 0} \frac{3 \ln \left(\frac{x}{\sin} x\right)}{x} ^ 2 = 3 {\lim}_{x \rightarrow 0} \frac{\ln \left(\frac{x}{\sin} x\right)}{x} ^ 2$

This limit has form $\frac{0}{0}$ so we use l'Hopital:

$= 3 {\lim}_{x \rightarrow 0} \frac{\sin \frac{x}{x} \left(\frac{\sin x - x \cos x}{{\sin}^{2} x}\right)}{2 x}$

$= \frac{3}{2} {\lim}_{x \rightarrow 0} \left(\frac{\sin x - x \cos x}{{x}^{2} \sin x}\right)$

This limit also has form $\frac{0}{0}$, so apply l'Hopital again.

$= \frac{3}{2} {\lim}_{x \rightarrow 0} \left(\frac{\cos x - \left(\cos x - x \sin x\right)}{2 x \sin x + {x}^{2} \cos x}\right)$

$= \frac{3}{2} {\lim}_{x \rightarrow 0} \left(\frac{\sin x}{2 \sin x + x \cos x}\right)$

Oh, No! It's STILL $\frac{0}{0}$!

Yes, but the derivative of the numerator is $\cos x$. So one more application of l'Hopital should get us an answer.

$= \frac{3}{2} {\lim}_{x \rightarrow 0} \left(\frac{\cos x}{2 \cos x + \left(\cos x - x \sin x\right)}\right)$

$= \frac{3}{2} \left(\frac{1}{2 + 1 + 0}\right) = \frac{3}{2} \left(\frac{1}{3}\right) = \frac{1}{2}$

Therefore,

${\lim}_{x \rightarrow 0} {\left(\frac{x}{\sin} x\right)}^{\frac{3}{x} ^ 2} = {\lim}_{x \rightarrow 0} {e}^{\frac{3}{x} ^ 2 \ln \left(\frac{x}{\sin} x\right)}$

$= {e}^{{\lim}_{x \rightarrow 0} \left(\frac{3}{x} ^ 2 \ln \left(\frac{x}{\sin} x\right)\right)}$

$= {e}^{\frac{1}{2}} = \sqrt{e}$

Nov 12, 2016

$\sqrt{e}$

Explanation:

sin x = x-x^3/(3!)+x^5/(5!)+ cdots and
sin x/x = 1-x^2/(3!)+x^4/(5!)+ cdots

both series are alternate and for $\left\mid x \right\mid \le 1$ we have

1-x^2/(3!) le sin x/x le 1-x^2/(3!)+x^4/(5!) and also

1/(1-x^2/(3!)+x^4/(5!)) le x/sin x le 1/(1-x^2/(3!) ) and also

(1/(1-x^2/(3!)+x^4/(5!)))^(3/x^2) le ( x/sin x)^(3/x^2) le (1/(1-x^2/(3!) ))^(3/x^2)

but

lim_(x->0) (1/(1-x^2/(3!) ))^(3/x^2) = lim_(x->0) (1/(1-x^2/(3!) ))^((3!)/x^2/2) and making y = x^2/(3!) we have

lim_(x->0) (1/(1-x^2/(3!) ))^((3!)/x^2/2) =sqrt(lim_(y->0)(1/(1-y))^(1/y)) = sqrt(e)

Analogously for

lim_(x->0) (1/(1-x^2/(3!)+x^4/(5!)))^(3/x^2) =sqrt(e)

so

${\lim}_{x \to 0} {\left(\frac{x}{\sin} x\right)}^{\frac{3}{x} ^ 2} = \sqrt{e}$