Question

evaluate the following limit.
lim _ { x
ightarrow infty } left( 1 + \frac { 1 } { x ^ { 3 } }
ight) ^ { x ^ { 2 } }
lim _ { x
ightarrow infty } left( 1 + \frac { 1 } { x ^ { 3 } }
ight) ^ { x ^ { 2 } } = square \text { (type an exact answer.) }

Explanation:

Step1: Use the formula $\lim_{t

ightarrow0}(1 + t)^{\frac{1}{t}}=e$
Let $t=\frac{1}{x^{3}}$, then as $x
ightarrow\infty$, $t
ightarrow0$. And $x^{2}=\frac{1}{t^{\frac{2}{3}}}$. So, $\lim_{x
ightarrow\infty}(1+\frac{1}{x^{3}})^{x^{2}}=\lim_{t
ightarrow0}(1 + t)^{\frac{1}{t^{\frac{2}{3}}}}$.

Step2: Rewrite the limit

We know that $(1 + t)^{\frac{1}{t^{\frac{2}{3}}}}=e^{\frac{\ln(1 + t)}{t^{\frac{2}{3}}}}$. Now, use the Taylor series expansion $\ln(1 + t)=t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots$. Then $\frac{\ln(1 + t)}{t^{\frac{2}{3}}}=\frac{t-\frac{t^{2}}{2}+\frac{t^{3}}{3}-\cdots}{t^{\frac{2}{3}}}=t^{\frac{1}{3}}-\frac{t^{\frac{4}{3}}}{2}+\frac{t^{\frac{7}{3}}}{3}-\cdots$.

Step3: Evaluate the limit of the exponent

As $t
ightarrow0$, $\lim_{t
ightarrow0}\frac{\ln(1 + t)}{t^{\frac{2}{3}}}=0$.

Step4: Evaluate the original limit

Since $\lim_{x
ightarrow\infty}(1+\frac{1}{x^{3}})^{x^{2}}=\lim_{t
ightarrow0}e^{\frac{\ln(1 + t)}{t^{\frac{2}{3}}}}$, and $\lim_{t
ightarrow0}\frac{\ln(1 + t)}{t^{\frac{2}{3}}}=0$, then $\lim_{x
ightarrow\infty}(1+\frac{1}{x^{3}})^{x^{2}}=e^{0}=1$.

Answer:

$1$