limit x->0 (e^x-1)/x — step-by-step solution

1) Let's evaluate the limit of

\frac{e^{x} - 1}{x}

when

x \to 0

ⓘ The base tends to a strictly positive number and the exponent to a finite number: substitute these limits by continuity

The limit of

e^{x}

1

when

x \to 0

ⓘ The limit of a sum of finite limits is the sum of the limits

The limit of

e^{x} - 1

1 - 1

when

x \to 0

Let's calculate:

1 - 1

ⓘ Start from

1

and move back

1

on the number line, down to zero:

1 - 1

0

0

For

\frac{e^{x} - 1}{x}

, the limit takes the indeterminate form 0/0 when

x

→

0

2) Let's apply L'Hôpital's rule (limit of a/b = limit of a'/b') by differentiating numerator and denominator:

3) Let's perform the differentiation:

\frac{d}{d x} (e^{x} - 1)

ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (e^{x} - 1)

\frac{d}{d x} (e^{x}) + 0

\frac{d}{d x} (e^{x}) + 0

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (e^{x}) + 0

\frac{d}{d x} (e^{x})

■The derivative of exp(x) is exp(x):

\frac{d}{d x} (e^{x})

e^{x}

e^{x}

4) Let's differentiate:

\frac{d}{d x} (x)

ⓘ The derivative of x is 1:

\frac{d}{d x} (x)

1

1

With the derivatives of the numerator and the denominator, we obtain a new fraction whose limit can be determined:

So the limit of

\frac{e^{x} - 1}{x}

1

when

x \to 0

▶Limit:

1