Question

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.
lim_{x
ightarrow0}\frac{e^{4x}-1}{2x^{2}+5x}
lim_{x
ightarrow0}\frac{e^{4x}-1}{2x^{2}+5x}=square \text{ (simplify your answer.)}

Explanation:

Step1: Check indeterminate form

When \(x
ightarrow0\), substitute \(x = 0\) into \(\frac{e^{4x}-1}{2x^{2}+5x}\). We get \(\frac{e^{4\times0}-1}{2\times0^{2}+5\times0}=\frac{1 - 1}{0}=\frac{0}{0}\), which is an indeterminate - form. So, we can apply L'Hopital's Rule.

Step2: Differentiate numerator and denominator

Differentiate the numerator \(y_1=e^{4x}-1\) using the chain - rule. The derivative \(y_1^\prime=\frac{d}{dx}(e^{4x}-1)=4e^{4x}\). Differentiate the denominator \(y_2 = 2x^{2}+5x\), and \(y_2^\prime=\frac{d}{dx}(2x^{2}+5x)=4x + 5\). So, the limit becomes \(\lim_{x
ightarrow0}\frac{4e^{4x}}{4x + 5}\).

Step3: Evaluate the new limit

Substitute \(x = 0\) into \(\frac{4e^{4x}}{4x + 5}\). We have \(\frac{4e^{4\times0}}{4\times0+5}=\frac{4\times1}{5}=\frac{4}{5}\).

Answer:

\(\frac{4}{5}\)