Single Variable Calculus, Chapter 2, 2.4, Section 2.4, Problem 42

Prove that the $\displaystyle \lim \limits_{x \to -3} \frac{1}{(x + 3)^4} = \infty$

Based from the definition, we let $M > 0$. So,

if $0 < | x - (-3) | < \delta$ then $\displaystyle \frac{1}{(x+3)^4} > M$

$0 < | x + 3 | < \delta$

But,


$
\begin{equation}
\begin{aligned}

& \frac{1}{(x + 3)^4} > M
&& \rightarrow (x + 3)^4 < \frac{1}{M}
&&& \rightarrow x + 3 < \sqrt[4]{\frac{1}{M}}\\

& \rightarrow x + 3 < \frac{1}{\sqrt[4]{M}}

\end{aligned}
\end{equation}
$


It shows that if we choose $\displaystyle \delta = \frac{1}{\sqrt[4]{M}}$, we will be able to prove that the $\displaystyle \lim \limits_{x \to -3} \frac{1}{(x + 3)^4} = \infty$