Limit of (e^3x - 1)/x at 0 Is 3

Make the denominator match the exponent

Put $u = 3x$. As $x \to 0$ so does $u$, and the quotient on the right becomes $\frac{e^{u}-1}{u}$, whose limit is $1$. The $3$ out front is untouched by any of this.

It is a derivative at a point

With $f(x) = e^{3x}$ you have $f(0) = 1$, so the quotient is $\frac{f(x)-f(0)}{x-0}$. That is the difference quotient for $f'(0)$, and $f'(x) = 3e^{3x}$.

Recognising the shape is the fastest route on a no-calculator section, and it explains why the exponent coefficient survives while everything else cancels.

The mistakes students make

Each of these misreads where the 3 sits in the expression.

• Mistaking this for $\frac{e^{x}-1}{x}$ and giving its limit, $1$. The $3$ in the exponent still has to be paid for.
• Reading the $3$ as a factor in front, as in $\frac{3e^{x}-1}{x}$. That numerator tends to $2$, not $0$, so the quotient has no finite limit at all.
• Cancelling the $x$ in the exponent against the $x$ in the denominator, as if $\frac{e^{3x}}{x} = e^{3}$, and reporting $e^{3}$. An $x$ sitting in an exponent is not a factor, so there is nothing there to cancel.