Finding $\lim_{x\to0}\tan(x)^{1/x}$

Question

I'm not sure how to evaluate this limit. $$ \lim_{x\to0^{+}} \tan(x)^{\frac{1}{x}} $$ I've attempted to use L'Hopital's rule, but I'm not sure if the indeterminate form (which I'll show below) from which I differentiate the numerator and denominator, is actually a true indeterminate form.

Here is my process. $$ \lim_{x\to0^{+}} \tan(x)^{\frac{1}{x}} $$ $$ \lim_{x\to0^{+}} e^{\ln(\tan(x)^{\frac{1}{x}})} $$ $$ \lim_{x\to0^{+}} e^{\frac{\ln(\tan(x))}{x}} $$ $$ \exp \lim_{x\to0^{+}} \frac{\ln(\tan(x))}{x} $$ From here, direct substitution yields $ -\infty $ on top and 0 on the bottom, so what I did in order to get it into indeterminate form (as previously stated, what I do here is somewhat odd so I'm not sure if this is correct): $$ \exp \lim_{x\to0^{+}} \frac{\ln(\tan(x))}{\frac{1}{\frac{1}{x}}} $$ At this point, direct substitution gives $ -\infty $ on top, and, on the bottom as soon as 0 is plugged in we get a $ \frac{1}{0} $. Now I know that division by zero is undefined, but the reason why I assumed that it was safe to treat it as infinity in the bottom was because, first of all, as $\frac{1}{x}$ approaches infinity it approaches $ 0 $, and additionally I've seen a similar technique of turning $ ux $ into $ \frac{u}{\frac{1}{x}} $ and using that to our advantage in limits like $ \lim_{x\to0^{+}} x^{x} $.

From this point on: $$ \exp \lim_{x\to0^{+}} \frac{\ln(\tan(x))}{\frac{1}{\frac{1}{x}}} $$ We apply L'Hopital's Rule to yield $$ \exp \lim_{x\to0^{+}} \frac{\frac{\sec^{2}(x)}{\tan(x)}}{\frac{-1}{x^{2}}}$$ $$ \exp \lim_{x\to0^{+}} \frac{-x^{2}}{\cos(x)\sin(x)} $$ And, once again, from here I got a $ \frac{0}{0} $ indeterminate form, and this I L'Hopitaled yet another time. $$ \exp \lim_{x\to0^{+}} \frac{-2x}{\cos^{2}(x)-\sin^{2}(x)} $$ Finally, direct substitution gives $ \exp(0) $, so $ 1 $. Now, the issue is, that according to almost every calculator like desmos and Wolfram Alpha, it is quite clear that the limit is actually $ 0 $, not what I have gotten.

I apologize for any blatant errors, over complications, and silly mistakes in my math and post. This is my first question on the Mathematics Stack Exchange. Could somebody please point out where I have made my error and how the limit is properly evaluated?

Thank You

Answer 1

Hint:

$$\lim_{x\to0^{+}} \tan(x)^{\frac{1}{x}}=\large \color{black}{\lim_{x \to 0^\mathtt{\text{+}}} e^{\frac{\ln{\left(\tan{\left(x \right)} \right)}}{x}}} = \color{black}{e^{\lim_{x \to 0^\mathtt{\text{+}}} \frac{\ln{\left(\tan{\left(x \right)} \right)}}{x}}}$$

The function decreases without a bound: $$\lim_{x \to 0^+} \frac{\ln{\left(\tan{\left(x \right)} \right)}}{x} = -\infty$$ Therefore,

$$\lim_{x \to 0^+} \tan^{\frac{1}{x}}{\left(x \right)} = 0$$

Answer 2

A more elementary approach without logarithms.

When $0<x<\pi/4,$ $0<\tan x<1$ and $\frac1x>1,$ so $$0<\tan(x)^{1/x}<\tan x\tag1$$ By the squeeze theorem, since $\tan x\to0,$ $\tan(x)^{1/x}\to0.$

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$(1)$ follows because when $0<u<1$ and $v>1$ then

$$0<u^v=u\cdot u^{v-1}<u$$

Answer 3

When you have :

$$ \tan(x)=x+O(x^3)$$

So

$$ \exp\left(\dfrac{\ln(\tan(x))}{x}\right)=\exp\left(\dfrac{\ln(x)+(\ln(1+O(x^2))}{x}\right) $$

Because $ \dfrac{\ln(x)}{x} \to -\infty$ as $x \to 0^+$

Your limit is $$0$$