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?

  • 1
    $\begingroup$ Why have you changed the title of the OP? $\endgroup$ – 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 $-1$.


A substitution can be helpful, as it transforms the expression into a rational function:

  • 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 $-1$.


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