## anonymous 4 years ago lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)

1. anonymous

2. dumbcow

$\lim_{? \rightarrow ?} \frac{t(e^{1/t^{3}}-1}{\sin(1/t)}$ is that right?

3. anonymous

lim t-> +oo

4. dumbcow

ok

5. anonymous

t*(eˆtˆ-3)-1

6. dumbcow

looks like you get 0/0 use L'hopitals rule and take derivative of top and bottom

7. dumbcow

so the minus 1 is on outside ?

8. anonymous

oh no in inside

9. anonymous

Let x = 1 / t, hence when t -> oo, x -> 0 hence the limit becomes Lim x -> 0 (e^x^3 - 1) / (sinx /x)

10. anonymous

lim x -> 0 sinx / x = 1 and the top becomes 0, hence the answer is 0

11. anonymous

did you get it?

12. anonymous

but the result of this limit is 0?

13. anonymous

yeah, that is acceptable

14. anonymous

it is a finite number

15. anonymous

did you get the process of reaching the answer

16. anonymous

aahhh ok

17. anonymous

you didi the substitution right? (sorry for my english!)

18. anonymous

yeah I did, I made x = 1 / t, so if t -> oo then x should be -> 0

19. dumbcow

mainaknag so t = 1/x right i don;t think you substituted correctly --> (e^x3 -1)/x*sinx

20. anonymous

I'm confuse

21. anonymous

yeah I am sorry

22. anonymous

ok let me try again, thank you dumbcow

23. dumbcow

but you are right, the limit is zero though :)

24. anonymous

hmm, I think so

25. dumbcow

not sure how to show it though...L'hopitals rule gets me nowhere

26. anonymous

you need to divide by x3 both numerator and denominator

27. anonymous

lim y -> 0 (e^y - 1) / y = 1, where y = x3 since x -> 0, y -> 0

28. anonymous

wait coming with full solution

29. 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 $\lim_{x \rightarrow 0} 1 / (1 / x) = \lim_{x \rightarrow 0} x = 0$

30. anonymous

just need to mention I substituted y = x3

31. anonymous

thank you