evaluate the following limit. use lhôpitals rule when it is convenient and applicable.\n lim_{x\rightarrow0}\…

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.\n lim_{x\rightarrow0}\frac{e^{4x}-1}{2x^{2}+5x} \n lim_{x\rightarrow0}\frac{e^{4x}-1}{2x^{2}+5x}=square \text{ (simplify your answer.)}

evaluate the following limit. use lhôpitals rule when it is convenient and applicable.\n lim_{x\rightarrow0}\frac{e^{4x}-1}{2x^{2}+5x} \n lim_{x\rightarrow0}\frac{e^{4x}-1}{2x^{2}+5x}=square \text{ (simplify your answer.)}

Answer

Explanation:

Step1: Check indeterminate form

When (x\rightarrow0), 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\rightarrow0}\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})