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OpenStudy (anonymous):
help with a limit problem ? Lol
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OpenStudy (anonymous):
@ganeshie8
OpenStudy (anonymous):
\[\lim_{x \rightarrow \infty} x^3/e^x/2\]
OpenStudy (anonymous):
x^3 / e^(x/2)
OpenStudy (anonymous):
sorry its supposed to look like the 2nd one
ganeshie8 (ganeshie8):
take L'hospital 3 times
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ganeshie8 (ganeshie8):
numerator reduces to a constant
however nothing much happens in the denominator
OpenStudy (anonymous):
so then the constant would be 6/e^x/2 / 8 ?
ganeshie8 (ganeshie8):
\[\large \lim\limits_{x\to\infty} \dfrac{x^3}{e^{x/2}} ~\stackrel{\text{LH 3 times}}{\leadsto\leadsto\leadsto} ~ \lim\limits_{x\to\infty} \dfrac{c}{e^{x/2}} = 0 \]
ganeshie8 (ganeshie8):
exponential function overtakes ANY polynomial
OpenStudy (anonymous):
what ?
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ganeshie8 (ganeshie8):
you're right, you get : 6/e^x/2 / 8
which is same as 48/e^(x/2)
ganeshie8 (ganeshie8):
next take the limit
OpenStudy (anonymous):
okay would it be 48/infinity ? lol
OpenStudy (anonymous):
so a constant over infinity equals 0 ????
OpenStudy (anonymous):
why did you leave @ganeshie8 still have unanswered questions :(
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OpenStudy (accessdenied):
Yep, a constant divided by an increasingly large number tends towards 0.
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