Question

lim t->+oo (t(eˆtˆ-3 - 1)/ sin(tˆ-1)

Answer 1

please help me

Answer 2

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

Answer 3

lim t-> +oo

Answer 4

ok

Answer 5

t*(eˆtˆ-3)-1

Answer 6

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

Answer 7

so the minus 1 is on outside ?

Answer 8

oh no in inside

Answer 9

Let x = 1 / t, hence when t -> oo, x -> 0 hence the limit becomes

Lim x -> 0 (e^x^3 - 1) / (sinx /x)

Answer 10

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

Answer 11

did you get it?

Answer 12

but the result of this limit is 0?

Answer 13

yeah, that is acceptable

Answer 14

it is a finite number

Answer 15

did you get the process of reaching the answer

Answer 16

aahhh ok

Answer 17

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

Answer 18

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

Answer 19

mainaknag
so t = 1/x right
i don;t think you substituted correctly

--> (e^x3 -1)/x*sinx

Answer 20

I'm confuse

Answer 21

yeah I am sorry

Answer 22

ok let me try again, thank you dumbcow

Answer 23

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

Answer 24

hmm, I think so

Answer 25

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

Answer 26

you need to divide by x3 both numerator and denominator

Answer 27

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

Answer 28

wait coming with full solution

Answer 29

\[\lim_{t \rightarrow \infty} t (e ^{1/t ^{3}} - 1) / \sin (1 / t)\]

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 \[\lim_{x \rightarrow 0} 1 / (1 / x) = \lim_{x \rightarrow 0} x = 0\]

Answer 30

just need to mention I substituted y = x3

Answer 31

thank you