Question

evaluate the following limits. if needed, enter oo for $infty$ and -oo for $-infty$, and dne if the limit does not exist.
$lim_{x
ightarrow4^{+}}\frac{2x}{(x - 4)^{4}}=$
$lim_{x
ightarrow4^{-}}\frac{2x}{(x - 4)^{4}}=$
$lim_{x
ightarrow4}\frac{2x}{(x - 4)^{4}}=$
score: n/a/0.667 n/a/0.667 n/a/0.667
had help available: video written example

Explanation:

Step1: Analyze right - hand limit

As $x\to4^{+}$, we have $2x\to2\times4 = 8$ and $(x - 4)^{4}>0$. Let $t=x - 4$, when $x\to4^{+}$, $t\to0^{+}$. The function becomes $\frac{2(t + 4)}{t^{4}}=\frac{2t+8}{t^{4}}$. As $t\to0^{+}$, the numerator approaches 8 and the denominator approaches 0 through positive values. So $\lim_{x\to4^{+}}\frac{2x}{(x - 4)^{4}}=\infty$.

Step2: Analyze left - hand limit

As $x\to4^{-}$, we have $2x\to2\times4 = 8$ and $(x - 4)^{4}>0$. Let $t=x - 4$, when $x\to4^{-}$, $t\to0^{-}$. But $(x - 4)^{4}=t^{4}>0$. The function $\frac{2x}{(x - 4)^{4}}=\frac{2(t + 4)}{t^{4}}$, as $t\to0^{-}$, the numerator approaches 8 and the denominator approaches 0 through positive values. So $\lim_{x\to4^{-}}\frac{2x}{(x - 4)^{4}}=\infty$.

Step3: Analyze overall limit

Since $\lim_{x\to4^{+}}\frac{2x}{(x - 4)^{4}}=\infty$ and $\lim_{x\to4^{-}}\frac{2x}{(x - 4)^{4}}=\infty$, then $\lim_{x\to4}\frac{2x}{(x - 4)^{4}}=\infty$.

Answer:

$\lim_{x\to4^{+}}\frac{2x}{(x - 4)^{4}}=\infty$
$\lim_{x\to4^{-}}\frac{2x}{(x - 4)^{4}}=\infty$
$\lim_{x\to4}\frac{2x}{(x - 4)^{4}}=\infty$