Question

find the following limit.
lim (8 + x^16 ln x)
x→0+
lim (8 + x^16 ln x) = (simplify your answer)
x→0+

Explanation:

Step1: Split the limit

By the sum - rule of limits $\lim_{x
ightarrow a}(f(x)+g(x))=\lim_{x
ightarrow a}f(x)+\lim_{x
ightarrow a}g(x)$, we have $\lim_{x
ightarrow0^{+}}(8 + x^{16}\ln x)=\lim_{x
ightarrow0^{+}}8+\lim_{x
ightarrow0^{+}}(x^{16}\ln x)$. Since $\lim_{x
ightarrow0^{+}}8 = 8$, we focus on $\lim_{x
ightarrow0^{+}}(x^{16}\ln x)$.

Step2: Rewrite the limit for L'Hopital's rule

We rewrite $x^{16}\ln x=\frac{\ln x}{x^{- 16}}$. As $x
ightarrow0^{+}$, we have the indeterminate form $\frac{-\infty}{\infty}$.

Step3: Apply L'Hopital's rule

L'Hopital's rule states that if $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}$ is in the indeterminate form $\frac{0}{0}$ or $\frac{\pm\infty}{\pm\infty}$, then $\lim_{x
ightarrow a}\frac{f(x)}{g(x)}=\lim_{x
ightarrow a}\frac{f'(x)}{g'(x)}$.
The derivative of $y = \ln x$ is $y'=\frac{1}{x}$, and the derivative of $y=x^{-16}$ is $y'=- 16x^{-17}$. So $\lim_{x
ightarrow0^{+}}\frac{\ln x}{x^{-16}}=\lim_{x
ightarrow0^{+}}\frac{\frac{1}{x}}{-16x^{-17}}=\lim_{x
ightarrow0^{+}}\frac{-x^{16}}{16}$.

Step4: Evaluate the limit

As $x
ightarrow0^{+}$, $\lim_{x
ightarrow0^{+}}\frac{-x^{16}}{16}=0$.

Step5: Combine the results

Since $\lim_{x
ightarrow0^{+}}(8 + x^{16}\ln x)=\lim_{x
ightarrow0^{+}}8+\lim_{x
ightarrow0^{+}}(x^{16}\ln x)$, and $\lim_{x
ightarrow0^{+}}8 = 8$ and $\lim_{x
ightarrow0^{+}}(x^{16}\ln x)=0$, we get $\lim_{x
ightarrow0^{+}}(8 + x^{16}\ln x)=8$.

Answer:

$8$