Evaluate the limit of (e^x - 1)/x as x approaches 0.

Step-by-step solution

1. 1

   Confirm the form is indeterminate. Substituting $x=0$ gives $\frac{e^{0}-1}{0}=\frac{0}{0}$, so direct substitution fails and some genuine work is needed. Recognising $0/0$ rather than assuming the answer is $0$ or undefined is the first decision.

2. 2

   Method 1: the Taylor series of $e^{x}$. Expanding about $0$,

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

   Dividing by $x$ cancels the leading term:

   $\frac{e^{x}-1}{x}=1+\frac{x}{2}+\frac{x^{2}}{6}+\cdots\longrightarrow 1\quad\text{as }x\to0.$

   The series also shows how the ratio approaches $1$: linearly, with slope $\tfrac12$.

3. 3

   Method 2: recognise a derivative. By definition,

   $f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}.$

   Taking $f(x)=e^{x}$, so $f(0)=1$, the given expression is the difference quotient for $e^{x}$ at $0$. Since $f'(x)=e^{x}$, the limit equals $f'(0)=e^{0}=1$. This derivation is the most fundamental of the three — the limit is essentially the statement that $e$ is the base whose exponential has slope $1$ at the origin.

4. 4

   Method 3: L'Hopital's rule. The $0/0$ form permits differentiating numerator and denominator separately:

   $\lim_{x\to0}\frac{e^{x}-1}{x}=\lim_{x\to0}\frac{e^{x}}{1}=e^{0}=1.$

   Use this one with care: it relies on already knowing $\frac{d}{dx}e^{x}=e^{x}$, which method 2 shows is the same fact as the limit itself, so quoting L'Hopital here is circular if the derivative was derived from this limit.

5. 5

   Check numerically. Evaluating the ratio: at $x=0.1$ it is $1.0517$, at $x=0.01$ it is $1.005$, at $x=-0.01$ it is $0.995$, and at $x=0.0001$ it is $1.00005$. The values close in on $1$ from both sides, matching the series prediction $1+\tfrac{x}{2}$ ✓.