Question

lim_{x\to - 4^{-}}\frac{1}{(x + 4)^{2}}=
lim_{x\to - 4^{+}}\frac{1}{(x + 4)^{2}}=
lim_{x\to - 4}\frac{1}{(x + 4)^{2}}=

x

\frac{1}{(x + 4)^{2}}

-5 | 1
-4.5 | 4
-4.1 | 100
-4.01 | 10,000
-4 | und
-3.99 | 10,000
-3.9 | 100
-3.5 | 4
-3 | 1

Explanation:

Step1: Analyze the limit as $x\to - 4^{-}$

We consider the function $y = \frac{1}{(x + 4)^2}$. As $x$ approaches $-4$ from the left - hand side, i.e., values of $x$ are slightly less than $-4$. Let $x=-4 - h$, where $h>0$ and $h\to0$. Then $(x + 4)^2=(-4 - h+4)^2=h^2$.

Step2: Find the limit value

Substituting into the function, we get $\lim_{h\to0}\frac{1}{h^2}$. As $h$ approaches $0$ from the positive side, $\frac{1}{h^2}$ approaches $+\infty$.

Step3: Analyze the limit as $x\to - 4^{+}$

Let $x=-4 + h$, where $h>0$ and $h\to0$. Then $(x + 4)^2=(-4 + h + 4)^2=h^2$. And $\lim_{h\to0}\frac{1}{h^2}=+\infty$.

Step4: Analyze the two - sided limit

Since $\lim_{x\to - 4^{-}}\frac{1}{(x + 4)^2}=\lim_{x\to - 4^{+}}\frac{1}{(x + 4)^2}=+\infty$, we have $\lim_{x\to - 4}\frac{1}{(x + 4)^2}=+\infty$.

Answer:

$+\infty$