arianna1453
  • arianna1453
FAN + MEDAL Calculus help!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
igigighjkl
  • igigighjkl
What is "Calculus"?
arianna1453
  • arianna1453
\[\lim_{x \rightarrow -\infty} x ^{4}e ^{x}\]
igigighjkl
  • igigighjkl
??? I can't do that! sorry. @mathmale

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Michele_Laino
  • Michele_Laino
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 \(g\)
arianna1453
  • arianna1453
I still dont understand it. @Michele_Laino
Michele_Laino
  • Michele_Laino
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
Michele_Laino
  • Michele_Laino
oops.. is still*...
Michele_Laino
  • Michele_Laino
sorry I meant indeterminated form, not undetermined form
arianna1453
  • arianna1453
See I keep getting \[\lim_{x \rightarrow -\infty} \left(\begin{matrix}-e ^{-x} \\ 24\end{matrix}\right)\]
arianna1453
  • arianna1453
@Michele_Laino
arianna1453
  • arianna1453
Nevermind I got the answer. It was 0.
Michele_Laino
  • Michele_Laino
ok! such limit is equal to zero, so, the limit of function \(f(x)\) is \(+\infty\)
Michele_Laino
  • Michele_Laino
sorry it is vice versa: we have: \[\large \begin{gathered} \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{{x^4}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{ - {e^{ - x}}}}{{4{x^3}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{12{x^2}}} = \mathop {\lim }\limits_{x \to - \infty } \frac{{{e^{ - x}}}}{{24}} = + \infty \hfill \\ \hfill \\ \Rightarrow \mathop {\lim }\limits_{x \to - \infty } {x^4}{e^x} = 0 \hfill \\ \end{gathered} \]
freckles
  • freckles
you don't need to consider the reciprocal function \[\lim_{x \rightarrow - \infty}x^4 e^{x}=\lim_{x \rightarrow - \infty} \frac{x^4}{e^{-x}}=...=\lim_{x \rightarrow - \infty} \frac{24}{e^{-x}}=0\]

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