Apply L'Hôpital's rule to evaluate indeterminate forms 0/0.

L'Hôpital's Rule — Evaluate lim(x→0) (eˣ - 1)/x

Question

Evaluate:

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

This limit appeared in JEE Main 2024 and is a foundational result — understanding it deeply will help you handle dozens of similar limits.

Solution — Step by Step

Substitute $x = 0$ directly: numerator gives $e^0 - 1 = 0$ , denominator gives $0$ . We get $\frac{0}{0}$ — a classic indeterminate form. Direct substitution fails, so L’Hôpital’s rule applies.

L’Hôpital’s rule says: if $\lim_{x \to a} f(x) = 0$ and $\lim_{x \to a} g(x) = 0$ (or both $\pm\infty$ ), then:

\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}

provided the right-hand limit exists. Both conditions check out here, so we’re good to proceed.

Differentiate top and bottom with respect to $x$ :

Numerator: $\frac{d}{dx}(e^x - 1) = e^x$
Denominator: $\frac{d}{dx}(x) = 1$

The limit becomes:

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

Now substitute $x = 0$ — no indeterminate form this time:

\frac{e^0}{1} = \frac{1}{1} = 1

Answer: $\displaystyle\lim_{x \to 0} \frac{e^x - 1}{x} = 1$

Why This Works

L’Hôpital’s rule works because near $x = 0$ , both $f(x) = e^x - 1$ and $g(x) = x$ shrink to zero at comparable rates. The ratio of their derivatives captures the relative speed at which they approach zero — and that ratio is what the limit actually measures.

Geometrically, $e^x - 1 \approx x$ for small $x$ (this is the first-order Taylor expansion of $e^x$ ). So $\frac{e^x - 1}{x} \approx \frac{x}{x} = 1$ near the origin. L’Hôpital’s rule is essentially a formal way of extracting this linear approximation.

This result — $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$ — is itself a standard limit worth memorising. You’ll use it as a building block for harder limits.

Alternative Method

We can use the Taylor series expansion of $e^x$ around $x = 0$ :

e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

So:

e^x - 1 = x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots

Dividing by $x$ :

\frac{e^x - 1}{x} = 1 + \frac{x}{2!} + \frac{x^2}{3!} + \cdots

As $x \to 0$ , every term except the first vanishes. So the limit is $\mathbf{1}$ .

The Taylor expansion method is often faster in JEE when L’Hôpital’s needs multiple rounds. If you see $\frac{e^x - 1}{x}$ , $\frac{\sin x}{x}$ , or $\frac{\ln(1+x)}{x}$ as sub-expressions inside a tougher limit, substitute the known result directly — don’t redo the derivation.

Common Mistake

Differentiating the fraction as a whole instead of numerator and denominator separately.

A very common error: students apply the quotient rule to $\frac{e^x - 1}{x}$ and differentiate the entire expression as one unit, getting $\frac{xe^x - (e^x - 1)}{x^2}$ . That is NOT what L’Hôpital’s rule asks for.

L’Hôpital’s rule says differentiate $f(x)$ and $g(x)$ independently, then take the ratio $\frac{f'(x)}{g'(x)}$ . This is fundamentally different from $\left(\frac{f}{g}\right)'$ .

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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