# What did I do wrong when taking $\displaystyle\lim_{x\to0^+}\tan(x)^{\tan(x)}$?

So I was bored, and decided to do some limits for fun, and after a while came up with this:$$\lim_{x\to0^+}\tan(x)^{\tan(x)}$$which I thought that I might be able to evaluate. Here is my attempt at doing so:$$\lim_{x\to0^+}\tan(x)^{\tan (x)}=\exp\left(\lim_{x\to0^+}\tan(x)\ln\tan(x)\right)$$and now since this is an indeterminate limit of the form $$0\cdot\infty$$, I can rewrite this as$$\exp\left(\lim_{x\to0^+}\dfrac{\ln\tan(x)}{1/\tan(x)}\right)$$and then take the derivative of both the top and bottom to get$$\exp\left(\lim_{x\to0^+}\dfrac{\csc(x)\sec(x)}{-\csc^2(x)}\right)=\exp\left(-\lim_{x\to0^+}\csc(x)\sec(x)\right)$$And now, rewriting $$\csc(x)\sec(x)$$ as $$1/(\sin(x)\cos (x))$$ and using the trigonometric identity$$2\sin(x)\cos(x)=\sin(2x)\implies\sin(x)\cos(x)=0.5\sin(2x)$$to rewrite our limit as$$\exp\left(-\lim_{x\to0^+}\csc(x)\sec(x)\right)=\exp\left(-\lim_{x\to0^+}2\csc(2x)\right)$$and then taking the limit from the right hand side gives us$$\exp(-\infty)=0$$which was weird, since I should have gotten 1, as confirmed by Desmos and Wolfram Alpha.

So my question is: What did I do wrong? I'm not sure what went wrong when taking this limit.

• why not substituting $y=\tan x$ ? Commented Jan 24 at 14:49
• Since $\tan(x)$ tends to $0$ , the limit should be the same as for $x^x$ for $x\to 0$ , which is $1$. Commented Jan 24 at 14:49
• I think you should have $\sec x / \csc x$ and not $\sec x \csc x$ Commented Jan 24 at 14:50
• @SineoftheTime Oh, wow, I must've simplified that wrong, giving me the incorrect answer Commented Jan 24 at 14:51
• @SineoftheTime "why not substituting $y=\tan x$?" I actually never thought of using substitution in a limit, I'll have to use that sometimes Commented Jan 24 at 14:52

Note that$$-\frac{\csc(x)\sec(x)}{\csc^2(x)}=-\frac{\sec(x)}{\csc(x)}=-\tan(x).$$So, you do get $$1$$.