Question

exercises 4.8 lhôpitals rule
score: 16/20 answered: 8/10
question 9
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evaluate the limit using lhôpitals rule
\\(\lim_{x\to 0} \frac{e^x + 3x - 1}{6x}\\)

Explanation:

Step1: Check the form of the limit

When \(x = 0\), \(\frac{e^{x}+3x - 1}{6x}=\frac{e^{0}+3\times0 - 1}{6\times0}=\frac{1 + 0-1}{0}=\frac{0}{0}\), which is an indeterminate form. So we can apply L'Hôpital's Rule.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator. The derivative of \(y = e^{x}+3x - 1\) is \(y^\prime=e^{x}+3\) (using the rules \((e^{x})^\prime=e^{x}\) and \((ax)^\prime=a\)). The derivative of \(y = 6x\) is \(y^\prime = 6\).
So \(\lim_{x
ightarrow0}\frac{e^{x}+3x - 1}{6x}=\lim_{x
ightarrow0}\frac{e^{x}+3}{6}\)

Step3: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{e^{x}+3}{6}\). We get \(\frac{e^{0}+3}{6}=\frac{1 + 3}{6}\)

Answer:

\(\frac{2}{3}\)