Limits of 1/(1 + 2^(1/x)) at 0 and infinity
Given the function $$ f(x) = \frac{1}{1+2^{1/x}}, $$ find $\displaystyle\lim_{x\to0^+} f(x)$, $\displaystyle\lim_{x\to0^-} f(x)$, and $\displaystyle\lim_{x\to+\infty} f(x)$.
Explanation of Method
The whole function is controlled by the inner exponent $\frac{1}{x}$, whose behavior changes drastically depending on how $x$ approaches its target: it tends to $0$, to $+\infty$, or to $-\infty$ in the three cases asked about. Tracking $2^{1/x}$ in each of these three regimes, then substituting into $\dfrac{1}{1+2^{1/x}}$, gives three different limiting values — this function has no single limit as $x\to0$, only different one-sided limits.
- Determine the behavior of $\frac{1}{x}$ in each of the three limiting cases.
- Determine the resulting behavior of $2^{1/x}$ in each case.
- Substitute into $f(x) = \dfrac{1}{1+2^{1/x}}$ and simplify.
Step 1 · Limit as $x \to +\infty$
As $x \to +\infty$, $\dfrac{1}{x} \to 0$, so:
$$ 2^{1/x} \to 2^0 = 1. $$
Substituting into $f(x)$:
$$ \lim_{x \to +\infty} f(x) = \frac{1}{1+1} = \frac{1}{2}. $$
Step 2 · Limit as $x \to 0^+$
As $x \to 0^+$, $\dfrac{1}{x} \to +\infty$, so $2^{1/x}$ grows without bound:
$$ 2^{1/x} \to +\infty. $$
Since the denominator $1+2^{1/x}$ grows without bound while the numerator stays fixed at $1$:
$$ \lim_{x \to 0^+} f(x) = \lim_{x\to0^+}\frac{1}{1+2^{1/x}} = 0. $$
Step 3 · Limit as $x \to 0^-$
As $x \to 0^-$, $\dfrac{1}{x} \to -\infty$, so $2^{1/x}$ decays to $0$:
$$ 2^{1/x} \to 2^{-\infty} = 0. $$
Substituting into $f(x)$:
$$ \lim_{x \to 0^-} f(x) = \frac{1}{1+0} = 1. $$
Summary Table
| Limit point | Behavior of $\frac1x$ | Behavior of $2^{1/x}$ | $\lim f(x)$ |
|---|---|---|---|
| $x\to+\infty$ | $\to 0$ | $\to 1$ | $\dfrac12$ |
| $x\to0^+$ | $\to+\infty$ | $\to+\infty$ | $0$ |
| $x\to0^-$ | $\to-\infty$ | $\to 0$ | $1$ |
Additional Notes
- Since $\lim_{x\to0^+}f(x)=0 \neq 1=\lim_{x\to0^-}f(x)$, the two-sided limit $\lim_{x\to0}f(x)$ does not exist — this function has a genuine jump discontinuity at $x=0$ (where it is undefined anyway, since $\frac1x$ is undefined there).
- By the same reasoning as Step 1, $\lim_{x\to-\infty}f(x)=\frac12$ as well, since $\frac1x\to0$ regardless of the sign from which $x\to\pm\infty$ — so $f$ has the same horizontal asymptote $y=\frac12$ on both ends.
- This function is a variant of the logistic (sigmoid) family; the steep jump between $0$ and $1$ as $x$ crosses $0$ is characteristic of functions built from $b^{1/x}$ with $b>1$.
Result
The three limits of $f(x) = \dfrac{1}{1+2^{1/x}}$ are: