Question

FAN + MEDAL
Calculus help!

Answer 1

What is "Calculus"?

Answer 2

\[\lim_{x \rightarrow -\infty} x ^{4}e ^{x}\]

Answer 3

??? I can't do that! sorry. @mathmale

Answer 4

if I call with \(f(x)=x^4e^x\) then I consider the inverse function:
\[\Large g\left( x \right) = \frac{1}{{f\left( x \right)}} = \frac{1}{{{x^4}{e^x}}} = \frac{{{e^{ - x}}}}{{{x^4}}}\]
please try to apply de L'Hopital rule to the function \(g\)

Answer 5

I still dont understand it. @Michele_Laino

Answer 6

If I apply the de l'Hopital rule, I get this:
\[\Large \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{{x^4}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - {e^{ - x}}}}{{4{x^3}}}\]
now the limit at the right side is again an undetermined form, so I can apply again the de L'Hopital Rule, and I get:
\[\Large \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{{x^4}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - {e^{ - x}}}}{{4{x^3}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{12{x^2}}}\]
which is stiil an undetermined form, please apply de L'Hopital rule

Answer 7

oops.. is still*...

Answer 8

sorry I meant indeterminated form, not undetermined form

Answer 9

See I keep getting \[\lim_{x \rightarrow -\infty} \left(\begin{matrix}-e ^{-x} \\ 24\end{matrix}\right)\]

Answer 10

@Michele_Laino

Answer 11

Nevermind I got the answer. It was 0.

Answer 12

ok! such limit is equal to zero, so, the limit of function \(f(x)\) is \(+\infty\)

Answer 13

sorry it is vice versa:
we have:
\[\large \begin{gathered}
\mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{{x^4}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - {e^{ - x}}}}{{4{x^3}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{12{x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{24}} = + \infty \hfill \\
\hfill \\
\Rightarrow \mathop {\lim }\limits_{x \to - \infty } {x^4}{e^x} = 0 \hfill \\
\end{gathered} \]

Answer 14

you don't need to consider the reciprocal function
\[\lim_{x \rightarrow - \infty}x^4 e^{x}=\lim_{x \rightarrow - \infty} \frac{x^4}{e^{-x}}=...=\lim_{x \rightarrow - \infty} \frac{24}{e^{-x}}=0\]