Question

q11 lhospitals rule

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

Explanation:

Step1: Check indeterminate form

When \(x = 0\), \(\frac{e^{x}-1}{x}=\frac{e^{0}-1}{0}=\frac{1 - 1}{0}=\frac{0}{0}\), which is an indeterminate form. So we can apply L'Hospital's Rule.

Step2: Apply L'Hospital's Rule

Differentiate the numerator and denominator. The derivative of \(y=e^{x}-1\) is \(y^\prime=e^{x}\), and the derivative of \(y = x\) is \(y^\prime=1\). Then \(\lim_{x
ightarrow0}\frac{e^{x}-1}{x}=\lim_{x
ightarrow0}\frac{(e^{x}-1)^\prime}{x^\prime}\).

Step3: Evaluate the limit

\(\lim_{x
ightarrow0}\frac{e^{x}}{1}\), substitute \(x = 0\) into \(\frac{e^{x}}{1}\), we get \(\frac{e^{0}}{1}=1\).

Answer:

\(1\)