Evaluate the following limit. Use l'Hopital's Rule when it is convenient and applicable, \( \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x} \) Use l'Hopital's Rule to rewnite the given limit se that it is not an indeterminate form, \( \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x}=\lim _{x \rightarrow 0}\left(\frac{e^{3 x}}{2}\right) \) Evaluate the \( \left.\operatorname{limit}^{2}\right) \) \( \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x}=1 \) (Type an exact answer.)

To evaluate the limit \( \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x} \), we first notice that both the numerator and denominator approach 0 as \( x \rightarrow 0 \), creating the indeterminate form \( \frac{0}{0} \). We can apply l'Hôpital's Rule, which states that we can take derivatives of the numerator and denominator. First, we differentiate the numerator: \[ \frac{d}{dx}(e^{3x}-1) = 3e^{3x} \] Next, we differentiate the denominator: \[ \frac{d}{dx}(x^2 + 7x) = 2x + 7 \] Now applying l'Hôpital's Rule, we have: \[ \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x} = \lim _{x \rightarrow 0} \frac{3e^{3x}}{2x + 7} \] Substituting \( x = 0 \) into the right-hand side gives: \[ = \frac{3e^{0}}{2(0) + 7} = \frac{3}{7} \] Therefore, the limit is: \[ \lim _{x \rightarrow 0} \frac{e^{3 x}-1}{x^{2}+7 x} = \frac{3}{7} \] This is the exact answer you're looking for!