Find the limit: lim x-> 0 (e^t - 1) / t^3?

Question

By substitution you get the indeterminate form 0/0, so I applied L'hospital's rule.

After applying the rule you get:

lim x->0 e^t / 3t^2 which yields 1/0, which leads me to believe I cannot apply l'hospital's rule a second time.

I know the limit is ∞, what am I missing here?

The initial question is supposed to be lim t->0 not x.

Answer 1

What you're missing is that 1/0 is a determinate form -- in all cases, if [x→c]lim f(x)=1 and [x→c]lim g(x)=0, then [x→c]lim |f(x)/g(x)| = ∞. Further, since in this case 3t² is always positive, then in fact [t→0]lim e^t/(3t²) = ∞. (if the denominator is always negative, then the limit would be -∞. Finally, if the denominator assumes both positive and negative values in every neighborhood of c, then the limit doesn't exist, because the function achieves both very large positive and very large negative values). As such, you only needed to apply L'hopital's rule once, not twice (and as you observed, if you applied it twice you would get the wrong answer).

Answer 2

The limit of e^t/3 as t goes to 0 is 1/3, while the limit of 1/t^2 as t goes to 0 is positive infinity, so the limit of the product is positive infinity since 1/3>0 (if we had -1/3, it would be negative infinity).

Answer 3

we may be able to stick with L'Hopital's rule to this. all of us comprehend that the reduce of x as x methods 0 is of route 0. The reduce of ln(x) as x methods 0 is negative infinity and on the grounds that it really is squared it turns into powerful infinity. we may be able to replace proper the following replace to a reduce of infinity/infinity by technique of creating proper the following adjustment at the same time as holding a similar meaning: lim x-->0 {[(lnx)^2]/[a million/x]} carry out L'Hopital's Rule: lim x-->0 {[2*lnx*(a million/x)]/[-a million/x^2]} Simplify: lim x-->0 {[-2*lnx]/[a million/x]} back we come across a reduce of infinity/infinity because the negative 2 cancels the negativity of ln, so we stick with L'Hopital's Rule back: lim x-->0 {[-2/x]/[-a million/x^2]} back we simplify: lim x-->0 {2*x} From the following we may be able to immediately change and locate the reduce: lim x-->0 {2*x} = 2*0 = 0

Answer 4

Represent the numerator using Taylor formula:
e^t = sum (t^k/k!) , k=0..infinity

the first summand will be killed by -1, and we obtain lim t->0 ((t+t^2/2+...)/t^3) = lim t->0 (1/t^2+ 1/2t+...) = infinity

Answer 5

actually l'hopital says that if f,g-->0 and f'/g' -->limit then f/g --> same limit

lim ( e^t-1)/t^3 = (-1+ 1+t + O(t^2) ) /t^3 = (1+O(t))/t^2 -->oo