How to calculate $\lim_{x \to - 1} \frac{2}{(x+1)^4}$ Calculate $$\lim_{x \to - 1} \frac{2}{(x+1)^4}$$
a) $0$
b) $\infty$
c) $-\infty$
d) $2$
I am able to see that it is equivalent the limit as $x$ approaches $-1$ of $\frac{2}{(x^2+2x+1)^2}$. 
I know that when doing limits to infinity this would be $0$ because the denominator has the highest exponent, but I am confused for $x$ approaches $-1$.
Is this a limit from the right and left kind of thing? Would the process be the same if I changed $-1$ to another number?
 A: Let's construct tables of values.
$$
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline
0 & 2\\
-0.9 & 20,000\\
-0.99 & 200,000,000\\
-0.999 & 2,000,000,000,000
\end{array}
\qquad
\begin{array}{c c}
x & f(x) = \dfrac{2}{(x + 1)^4}\\ \hline 
-2 & 2\\
-1.1 & 20,000\\
-1.01 & 200,000,000\\
-1.001 & 2,000,000,000,000
\end{array}
$$
The values of $x$ have been chosen so that the $x$-coordinates of the points in the table at left are, respectively, $1$, $0.1$, $0.01$, and $0.001$ units greater than $-1$ and so that the $x$-coordinates of the points in the table at right are, respectively, $1$, $0.1$, $0.01$, and $0.001$ units less than $-1$.
Examining the values in the table suggests that as $x \to -1$, $f(x)$ increases without bound.
Let $M > 0$.  We will show that we can find $x$ sufficiently close to $-1$ such that $f(x) > M$.
\begin{align*}
f(x) & > M\\
\frac{2}{(x + 1)^4} & > M\\
2 & > M(x + 1)^4\\
\frac{2}{M} & > (x + 1)^4\\
\sqrt[4]{\frac{2}{M}} & > |x + 1| 
\end{align*}
Since the steps are reversible (the solution set of each step of the inequality is the same), the final inequality is equivalent to the initial one.  Since $|x + 1| = |x - (-1)|$ is the distance of $x$ from $-1$, we may conclude that whenever the distance of $x$ from $-1$ is smaller than $\sqrt[4]{\frac{2}{M}}$, then $f(x) > M$.  Since $M$ is arbitrary, $f(x)$ grows larger than any finite number $M$ as $x \to -1$.  Therefore,
$$\lim_{x \to -1} f(x) = \lim_{x \to - 1} \frac{2}{(x + 1)^4} = \infty$$
The function
$$f(x) = \frac{2}{(x + 1)^4}$$
is a rational function.  Its implicit domain is the set of all real numbers except those where the denominator is zero, which is the set of all real numbers except $-1$.  Rational functions are continuous on their domains.  Since a function is equal to its limit at a point of continuity, if $x_0 \in (-\infty, -1) \cup (-1, \infty)$, then
$$\lim_{x \to x_0} f(x) = f(x_0)$$
which means we can simply substitute the value of $x_0$ into the function to find the limit at $x = x_0$.  For example,
$$\lim_{x \to 0} f(x) = f(0) = 2$$
This is not true at $x = -1$ since it lies outside the domain of the function.  Consequently, we must check what value, if any, the function approaches as $x \to - 1$.
A: It approaches infinity. The reason for this is because as x approaches -1 in the denominator, the denominator gets closer and closer to 0 (but is always greater than 0). This means the entire function becomes something like 2/0. To think about this consider 2/1, 2/0.1, 2/0.01, etc and notice that it keeps getting larger. 
A: Consider this:
$$ x\to-1\qquad\Leftrightarrow\qquad (x+1)\to 0$$
$$ \lim_{x\to-1}\frac{2}{(x+1)^4} = \lim_{(x+1)\to 0} \frac{2}{(x+1)^4} = \lim_{y\to 0} \frac{2}{y^4}$$
A: Consider the easier function $\frac{1}{(x+1)^{2}}$. What would be it's value if $x$ approach -1 ?
A: $$\lim_{x \to - 1} \frac{2}{(x+1)^4} = \frac{\lim_{x \to -1}(2)}{\lim_{x \to -1} (x+1)^{4}} = \frac{2}{(+)0} = +\infty$$
