# Limit of $5\cdot (\tan x)^{\sin x}$ at $0^+$, and indeterminate forms

I am looking for the right hand limit $$\lim_{x\to 0^+} 5\cdot (\tan x)^{\sin x}$$

I realize that I need to apply l'Hopital's rule but I'm having trouble getting the indeterminate form.

• The 5 upfront is of no use here. Write $y=(tanx)^{sinx}$ and take $ln$ on both sides. Then consider the limit on $(sinx)ln(tanx)$ and go from there towards LHospital Rule – imranfat Mar 8 '17 at 19:17

First, side remark: this is well-defined since $\sin x > 0$ for $x>0$ sufficiently small, so we do not have to deal with powers of a negative number.

Also, no, you don't need to use L'Hopital's rule -- below is a detailed approach which does not use it at all,instead relying on 3 elementary standard limits.

Rewrite it as follows, using the more intuitive and manageable exponential form:

$$5(\tan x)^{\sin x} = 5e^{\sin x \cdot \ln(\tan x)}$$ By continuity of $\exp$, it is enough to find the limit of the exponent. (Namely, if $\lim_{x\to 0^+} \sin x \cdot \ln(\tan x) = \ell$, then $\lim_{x\to 0^+} 5 e^{\sin x \cdot \ln(\tan x)} = 5 e^\ell$).

For $x>0$, \begin{align} \sin x \cdot \ln(\tan x) &= \frac{\sin x}{x}\cdot x \ln(\tan x) = \frac{\sin x}{x}\cdot x \ln(\frac{\tan x}{x}\cdot x)\\ &= \frac{\sin x}{x}\cdot \left( x \ln\frac{\tan x}{x}+ x\ln x \right) \end{align} where our goal was to make more manageable limits "appear."

In particular: \begin{align} \frac{\sin x}{x} &= \frac{\sin x-\sin 0}{x-0} \xrightarrow[x\to 0^+]{} \sin' 0 = \cos 0 =1 \tag{1}\\ \frac{\tan x}{x} &= \frac{\tan x-\tan 0}{x-0} \xrightarrow[x\to 0^+]{} \tan' 0 = \frac{1}{\cos^2 0} =1 \tag{2}\\ x\ln x &\xrightarrow[x\to 0^+]{} 0 \tag{3}\\ \end{align} and therefore we have no indeterminate form anymore, as \begin{align} \sin x \cdot \ln(\tan x) &= \frac{\sin x}{x}\cdot \left( x \ln\frac{\tan x}{x}+ x\ln x \right) \xrightarrow[x\to 0^+]{(1), (2), (3)} 1\cdot \left(0\cdot \ln 1+0\right) = 0 \tag{\dagger} \end{align} and, finally, $$5(\tan x)^{\sin x} = 5e^{\sin x \cdot \ln(\tan x)} \xrightarrow[x\to 0^+]{(\dagger)} 5e^0 = \boxed{5}.$$

If you want to use L'Hopital's Rule, here's one way you can find a familiar indeterminant form.

First note that $u = e^{\ln u}$ since $\ln$ is the inverse of natural exponentiation. So we can write \begin{align} 5\left(\tan x \right)^{\sin x} &= 5e^{\ln\left[(\tan x)^{\sin x} \right]} \\ &= 5e^{\sin x \ln(\tan x)} \end{align} Now, \begin{align} \sin x \ln(\tan x) &= \frac{\ln(\tan x)}{\csc x} \sim \frac{-\infty}{\infty} \ \leftarrow \textrm{Indeterminant form} \end{align} as $x \to 0^+$. By L'Hopital's Rule: \begin{align} \lim_{x \to 0^+} \frac{\ln(\tan x)}{\csc x} &= \lim_{x \to 0^+} \frac{\frac{1}{\tan x} \cdot \sec^2 x}{-\csc x \cot x} \ \ \ \ \textrm{(Derivative of top and bottom)} \\ &= -\lim_{x \to 0^+} \frac{\sec^2 x}{\csc x \cot x \tan x} \\ &= -\lim_{x \to 0^+} \frac{\sec^2 x}{\csc x} \ \ \ \ \textrm{(} \cot \textrm{ and } \tan \textrm{ cancel)} \\ &= -\lim_{x \to 0^+} \frac{1/\cos^2 x}{1/\sin x} \\ &= -\lim_{x \to 0^+} \frac{\sin x}{\cos^2 x} \\ &= -\frac{\sin(0)}{\cos^2(0)} \\ &= 0 \end{align} Hence \begin{align} \lim_{x \to 0^+} 5e^{\sin x \ln(\tan x)} = 5e^0 = 5 \end{align}

Much shorter: with equivalents:

$\tan x,\; \sin x\sim_{0}x$ and $\tan x >0$ if $x>0$ in a neighbourhood of $0$, hence $$\ln\bigl((\tan x)^{\sin x}\bigr)=\sin x\ln(\tan x)\sim_{0^+}x\ln x\to 0.$$

This solution does not require L'Hopital. However, it does require the following daunting identity. To learn more about it (trust me, once you've used it once, you'll never want to forget it), see this.

Recall the identity: $$\lim\limits_{x \to a}{\phi(x)^{\psi(x)}} = e^{\lim\limits_{x \to a}{[(\phi(x)-1)\psi(x)}]}$$

That being said: $$\lim\limits_{x \to 0}{5 \tan x^{\sin x}} = 5 \lim\limits_{x \to 0}{\tan x^{\sin x}} = 5 e^{\lim\limits_{x \to 0}{(\tan x - 1)\sin x}} = 5 e^0 = 5$$

$$\lim\limits_{x \to 0}{5 \tan x^{\sin x}} = 5$$