anonymous
  • anonymous
lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
please help me
dumbcow
  • dumbcow
\[\lim_{? \rightarrow ?} \frac{t(e^{1/t^{3}}-1}{\sin(1/t)}\] is that right?
anonymous
  • anonymous
lim t-> +oo

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

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