# Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises - Page 307: 16

The solution is $$\lim_{x\to 0}\frac{e^x-1}{x^2+3x}=\frac{1}{3}.$$

#### Work Step by Step

We will apply the L'Hopital's rule to solve this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to 0}\frac{e^x-1}{x^2+3x}=\left[\frac{0}{0}\right][\text{LH}]=\lim_{x\to 0}\frac{(e^x-1)'}{(x^2+3x)'}=\lim_{x\to0}\frac{e^x}{2x+3}=\left[\frac{e^0}{2\cdot0+3}\right]=\frac{1}{3}.$$

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