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anonymous
 4 years ago
lim t>+oo (t(eˆtˆ3  1)/ sin(tˆ1)
anonymous
 4 years ago
lim t>+oo (t(eˆtˆ3  1)/ sin(tˆ1)

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dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{? \rightarrow ?} \frac{t(e^{1/t^{3}}1}{\sin(1/t)}\] is that right?

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0looks like you get 0/0 use L'hopitals rule and take derivative of top and bottom

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0so the minus 1 is on outside ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Let x = 1 / t, hence when t > oo, x > 0 hence the limit becomes Lim x > 0 (e^x^3  1) / (sinx /x)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lim x > 0 sinx / x = 1 and the top becomes 0, hence the answer is 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but the result of this limit is 0?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah, that is acceptable

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is a finite number

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0did you get the process of reaching the answer

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you didi the substitution right? (sorry for my english!)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0yeah I did, I made x = 1 / t, so if t > oo then x should be > 0

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0mainaknag so t = 1/x right i don;t think you substituted correctly > (e^x3 1)/x*sinx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok let me try again, thank you dumbcow

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0but you are right, the limit is zero though :)

dumbcow
 4 years ago
Best ResponseYou've already chosen the best response.0not sure how to show it though...L'hopitals rule gets me nowhere

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0you need to divide by x3 both numerator and denominator

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lim y > 0 (e^y  1) / y = 1, where y = x3 since x > 0, y > 0

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wait coming with full solution

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\lim_{t \rightarrow \infty} t (e ^{1/t ^{3}}  1) / \sin (1 / t)\] Let x > 1 / t, then x > 0 And the limit becomes \[\lim_{x \rightarrow 0} (e ^{x ^{3}}  1) / x.sinx = \lim_{x \rightarrow 0} \lim_{y \rightarrow 0}(1 /y).(e ^{y}  1) / (1 / x) . (x / x) . (Sinx / x)\] the numerator > 1, lim x > 0 sinx / x = 1, lim x > 0 x / x = 1 Hence the limit becomes \[\lim_{x \rightarrow 0} 1 / (1 / x) = \lim_{x \rightarrow 0} x = 0\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just need to mention I substituted y = x3
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