find the limit. use lhospitals rule where appropriate. if there is a more elementary method, consider using…

Explanation:

Step1: Check indeterminate form

When $x\rightarrow0$, we have $\lim_{x\rightarrow0}\frac{x7^{x}}{7^{x}-1}=\frac{0\times7^{0}}{7^{0}-1}=\frac{0}{0}$, which is an indeterminate - form. So we can apply L'Hopital's Rule.

Step2: Differentiate numerator and denominator

The derivative of the numerator $y = x7^{x}$ using the product - rule $(uv)^\prime=u^\prime v + uv^\prime$, where $u = x$ and $v = 7^{x}$. The derivative of $u=x$ is $u^\prime = 1$, and the derivative of $v = 7^{x}$ is $v^\prime=7^{x}\ln7$. So $(x7^{x})^\prime=7^{x}+x7^{x}\ln7$. The derivative of the denominator $y = 7^{x}-1$ is $y^\prime=7^{x}\ln7$.

Step3: Find the new limit

Now we find $\lim_{x\rightarrow0}\frac{7^{x}+x7^{x}\ln7}{7^{x}\ln7}$. Substitute $x = 0$ into the new function: $\frac{7^{0}+0\times7^{0}\ln7}{7^{0}\ln7}=\frac{1 + 0}{\ln7}=\frac{1}{\ln7}$.

Answer:

$\frac{1}{\ln7}$