L’Hôpital’s rule, question 26

A student evaluates the limit below as shown. Find the first line that is wrong.

\lim_{x \to 0} \frac{e^{2x} - 1}{x}

$\text{At } x = 0 \text{ the form is } \tfrac{0}{0}, \quad \text{so l'Hôpital's rule applies.}$
Check the form first.
$\lim_{x \to 0} \frac{e^{2x} - 1}{x} = \lim_{x \to 0} \frac{2e^{2x}\cdot x - \bigl(e^{2x} - 1\bigr)\cdot 1}{x^{2}}$
Differentiate the fraction with the quotient rule.
$= \lim_{x \to 0} \frac{2xe^{2x} - e^{2x} + 1}{x^{2}}$
Tidy the numerator.
$\text{This is } \tfrac{0}{0} \text{ again,} \quad \text{so apply the rule once more.}$
Keep going until the form is determinate.

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