To evaluate the limit
[tex]\[
\lim_{x \rightarrow 0} \frac{e^{2x}-1}{3x^2+5x},
\][/tex]
we notice that substituting [tex]\( x = 0 \)[/tex] directly gives us an indeterminate form of type [tex]\(\frac{0}{0}\)[/tex]. Thus, it is appropriate to use l'Hôpital's Rule.
Step 1: Differentiate the numerator and the denominator
According to l'Hôpital's Rule, we differentiate the numerator and the denominator separately:
- Derivative of the numerator [tex]\(e^{2x} - 1\)[/tex]:
[tex]\[ \frac{d}{dx} [e^{2x} - 1] = 2e^{2x} \][/tex]
- Derivative of the denominator [tex]\(3x^2 + 5x\)[/tex]:
[tex]\[ \frac{d}{dx} [3x^2 + 5x] = 6x + 5 \][/tex]
Step 2: Form a new limit using the derivatives
Having differentiated both, we now form the new limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{2e^{2x}}{6x + 5}
\][/tex]
Step 3: Evaluate the new limit by direct substitution
With the new expression, we evaluate the limit as [tex]\( x \)[/tex] approaches 0:
[tex]\[
\lim_{x \rightarrow 0} \frac{2e^{2x}}{6x + 5}
\][/tex]
Substituting [tex]\( x = 0 \)[/tex] in the new expression:
[tex]\[
\frac{2e^{2 \cdot 0}}{6 \cdot 0 + 5} = \frac{2 \cdot e^{0}}{5} = \frac{2 \cdot 1}{5} = \frac{2}{5}
\][/tex]
Thus, the limit is:
[tex]\[
\lim_{x \rightarrow 0} \frac{e^{2x} - 1}{3x^2 + 5x} = \frac{2}{5}
\][/tex]
Therefore, the evaluated limit is [tex]\(\boxed{\frac{2}{5}}\)[/tex].
