Find lim as x to + infinity (e^x-2^x)/x

Find $\displaystyle\lim_{x\to+\infty}\frac{e^{x}-2^{x}}{x}$.

Step 1: Compare the growth rates

As $x\to+\infty$ all three of $e^{x}$, $2^{x}$ and $x$ tend to $+\infty$. Since $e\gt 2$, the exponential $e^{x}$ grows strictly faster than $2^{x}$, so the numerator is dominated by $e^{x}$ and is positive for large $x$.

Step 2: Factor out $e^{x}$

\[ \frac{e^{x}-2^{x}}{x} = \frac{e^{x}\left(1-\left(\dfrac{2}{e}\right)^{x}\right)}{x} = \frac{e^{x}}{x}\cdot\left(1-\left(\frac{2}{e}\right)^{x}\right). \tag{1} \]

Step 3: Analyse each factor

Because $\dfrac{2}{e} = 0.7358\lt 1$, the power $\left(\dfrac{2}{e}\right)^{x}\to0$, so the bracket tends to $1$: $1-\left(\frac{2}{e}\right)^{x} \longrightarrow 1$.

The other factor is unbounded: one application of l'Hôpital's rule to $\dfrac{e^{x}}{x}$, a form $\dfrac{\infty}{\infty}$, gives $\dfrac{e^{x}}{1}\to+\infty$.

Step 4: Conclude

A factor tending to $+\infty$ multiplied by one tending to $1$ gives, from (1), \[\frac{e^{x}-2^{x}}{x} = \frac{e^{x}}{x}\cdot\left(1-\left(\frac{2}{e}\right)^{x}\right) \longrightarrow \bigl(+\infty\bigr)\cdot1 = +\infty.\]

Step 5: Verification

\[x = 5:\ 23.283; \qquad x = 10:\ 2100.25; \qquad x = 20:\ 2.42\cdot10^{7} \;\;\checkmark\]

\[\left(\tfrac{2}{e}\right)^{20} = 2.3\cdot10^{-3},\ \text{so at } x = 20 \text{ the bracket is already } 0.9977 \;\;\checkmark\]

\[\text{Twice by l'H\^opital: } \frac{e^{x}-2^{x}\ln2}{1} \to +\infty \ \text{since } e^{x} \text{ dominates } 2^{x} \;\;\checkmark\]