I'm trying to compute this limit without the use of L'Hopital's rule:

$$\lim_{x \to 0^{+}} \frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}$$

I've been trying to multiply by the lcd and doing other creative stuff... anyone have any suggestions on theorems or techniques?

• Why have you changed the title of the OP? – Paras Khosla Apr 2 '19 at 7:24

Write the limit as $$\lim_{x\to 0+}\frac{1+4^{-2/x}}{-1+4^{-2/x}}$$ and use the fact that $$\lim_{x\to 0+}\frac{-2}{x}=-\infty.$$ to find that the limit equals $$-1$$.
• Set $$y=4^{\frac{1}{x}}$$ and consider $$y \to +\infty$$
$$\begin{eqnarray*} \frac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}} & \stackrel{y=4^{\frac{1}{x}}}{=} & \frac{\frac{1}{y}+y}{\frac{1}{y}-y} \\ & = & \frac{\frac{1}{y^2}+1}{\frac{1}{y^2}-1} \\ & \stackrel{y \to +\infty}{\longrightarrow} & \frac{0+1}{0-1} = -1 \end{eqnarray*}$$
$$\lim_{x\to 0^+}\dfrac{4^{-1/x}+4^{1/x}}{4^{-1/x}-4^{1/x}}=\lim_{x\to 0^+}\dfrac{4^{-2/x}+1}{4^{-2/x}-1}$$
Clearly as $$x\to 0^+$$, $$2/x\to \infty$$. Since the power of $$4$$ is $$-2/x$$, it must go to $$0$$. Effectively we have $$\frac{0+1}{0-1}=-1$$. Hence the required limit is $$-1$$.