Show that the limit as x approaches zero for $\frac{2^{1/x} - 2^{-1/x}}{2^{1/x} + 2^{-1/x}}$ does not exist This problem was in one of the first chapters of a calculus text, so how would you go about solving this without applying L'Hôpital's rule?
I attempted factoring out $2^{1/x}$, as well as using u substitution for $2^{1/x}$.
By graphing, I can see that the left-side limit does not equal the right-side limit, but how else can I demonstrate this?
 A: We can simplify the problem as follows:
$$\mathrm f(x) := \frac{2^{1/x}-2^{-1/x}}{2^{1/x}+2^{-1/x}} \equiv \frac{2^{2/x}-1}{2^{2/x}+1}
\equiv \frac{4^{1/x}-1}{4^{1/x}+1} \equiv \frac{1-4^{-1/x}}{1+4^{-1/x}}$$
All we need to do is consider the limits of $4^{1/x}$ and $4^{-1/x}$ as $x$ tends to zero.


*

*If $x<0$ and $x \to 0$ then $1/x \to - \infty$ meaning that $4^{1/x} \to 0$ and so $$\mathrm f(x) \equiv \frac{4^{1/x}-1}{4^{1/x}+1} \to \frac{0-1}{0+1} = -1$$

*If $x>0$ and $x \to 0$ then $-1/x \to -\infty$ meaning that  $4^{-1/x} \to 0$ and so $$\mathrm f(x) \equiv \frac{1-4^{-1/x}}{1+4^{-1/x}} \to \frac{1-0}{1+0} = 1$$


Since the left- and right-hand limits are different, the limit is not well-defined.
A: Quite simple with equivalents:
$$2^{1/x}\pm2^{-1/x}\sim_{0^+} 2^{1/x},\enspace\text{hence}\quad\frac{2^{1/x}-2^{-1/x}}{2^{1/x}+2^{-1/x}}\sim_{0^+}\frac{2^{1/x}}{2^{1/x}}=1,$$
while
$$2^{1/x}\pm2^{-1/x}\sim_{0^-}\pm 2^{-1/x},\enspace\text{hence}\quad\frac{2^{1/x}-2^{-1/x}}{2^{1/x}+2^{-1/x}}\sim_{0^-}\frac{-2^{-1/x}}{2^{-1/x}}=-1.$$
