given the following limit:
$$ \lim _ {x \to 0 } \left (\frac{\tan x } {x} \right ) ^{1/x^2}\;, $$
is there any simple way to calculate it ?
I have tried writing it as $e^ {\ln (\dots)} $ , but it doesn't give me anything [ and I did l'Hospital on the limit I received… ]
can someone help me with this?
thanks a lot !
Best Answer
If you know Taylor expansion, you know that $$\tan x = x + \frac{x^3}{3}+ \mathcal{O}(x^5)$$ where the big-Oh denotes a term which scales like $x^5$ for $x\to 0$. Thus, $$\frac{\tan x}{x} = 1 + \frac{x^2}{3} + \mathcal{O}(x^4).$$ The expansion of the logarithm around $1$ reads $$ \ln (1+y) = y + \mathcal{O}(y^2).$$ Letting $1+y=\tan x/x =1+ x^2/3 + \mathcal{O}(x^4)$, we obtain $$ \ln \left(\frac{\tan x}x \right) = \frac{x^2}3 + \mathcal{O}(x^4).$$ Now, $$ \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13 + \mathcal{O}(x^2).$$ And thus $$\lim_{x\to 0} \frac1 {x^2} \ln \left(\frac{\tan x}x \right) = \frac13.$$
With that you can easily show that $$\lim_{x\to 0} \left(\frac{\tan x}x\right) ^{x^{-2}} = \sqrt[3]{e}.$$