lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)

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lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)

Mathematics
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please help me
\[\lim_{? \rightarrow ?} \frac{t(e^{1/t^{3}}-1}{\sin(1/t)}\] is that right?
lim t-> +oo

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Other answers:

ok
t*(eˆtˆ-3)-1
looks like you get 0/0 use L'hopitals rule and take derivative of top and bottom
so the minus 1 is on outside ?
oh no in inside
Let x = 1 / t, hence when t -> oo, x -> 0 hence the limit becomes Lim x -> 0 (e^x^3 - 1) / (sinx /x)
lim x -> 0 sinx / x = 1 and the top becomes 0, hence the answer is 0
did you get it?
but the result of this limit is 0?
yeah, that is acceptable
it is a finite number
did you get the process of reaching the answer
aahhh ok
you didi the substitution right? (sorry for my english!)
yeah I did, I made x = 1 / t, so if t -> oo then x should be -> 0
mainaknag so t = 1/x right i don;t think you substituted correctly --> (e^x3 -1)/x*sinx
I'm confuse
yeah I am sorry
ok let me try again, thank you dumbcow
but you are right, the limit is zero though :)
hmm, I think so
not sure how to show it though...L'hopitals rule gets me nowhere
you need to divide by x3 both numerator and denominator
lim y -> 0 (e^y - 1) / y = 1, where y = x3 since x -> 0, y -> 0
wait coming with full solution
\[\lim_{t \rightarrow \infty} t (e ^{1/t ^{3}} - 1) / \sin (1 / t)\] 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 \[\lim_{x \rightarrow 0} 1 / (1 / x) = \lim_{x \rightarrow 0} x = 0\]
just need to mention I substituted y = x3
thank you

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