Q.Evaluate $\displaystyle\lim_{x\to0}\frac{e^{5x}-1}{x}$.

This matches $\dfrac{e^{kx}-1}{x}$ with $k=5$, which equals $k$ directly.\n> [!TLDR]\n> Scaled exponential-limit formula with $k=5$.\n …

Direct substitution gives $\dfrac{e^0-1}{0}=\dfrac{1-1}{0}=\dfrac00$ — indeterminate, so the standard exponential limit (§5) is used.

Substitute $u=5x$ (so $u\to0$ as $x\to0$, and $x=u/5$):

$\lim_{x\to0}\frac{e^{5x}-1}{x} = \lim_{u\to0}\frac{e^{u}-1}{u/5} = 5\cdot\lim_{u\to0}\frac{e^u-1}{u} = 5(1) = 5.$ …

Numeric verification (shown above): plugging in $x=0.001$ gives a value close to $5$, and smaller $x$ values converge even closer, inde …

A common slip is writing the scaled result as $e^5$ (confusing the coefficient $k$ in $\lim\frac{e^{kx}-1}{x}=k$ with an exponent) — the standard exponenti …