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

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

• The limit is $0^\infty$ which is zero. It's not a 'indeterminate' form Commented Feb 11, 2023 at 3:56
• @Dqrksun Oh thank you, I see. What a foolish mistake I have made... I guess this is what math teachers mean when they say that students will often go gung-ho on using L'Hopitals rule on every limit, even if it is not applicable. Commented Feb 11, 2023 at 4:21
• When $x\to 0^+$ we can assume $0<\tan x <1$ and $1/x>1$ so that we get $(\tan x) ^{1/x}<\tan x$. You can now apply squeeze. One should think of L Hospital 's Rule when every other method appears to be not working. Use it at last and not at first. Commented Feb 11, 2023 at 5:14
• @ParamanandSingh Thank you for the advice sir. Commented Feb 11, 2023 at 6:02

A more elementary approach without logarithms.

When $$0 $$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.$$

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

$$0

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,

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

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 $$0$$

• Why does $\ln x / x$ tend to $0$ as $x \to 0^+$? I think $\ln x \to -\infty$.
– Chia
Commented Feb 11, 2023 at 3:56
• I corrected it thanks
– EDX
Commented Feb 11, 2023 at 4:01