Tricky Limit question

Question

What is

$$\lim_{t \rightarrow 0} \frac{e^{t^3}-1-t^3}{\sin(t^2)-t^2}?$$

I used Hopital's rule, but it kept getting more complicated!

Maybe try taylor series expansions – AlexM Apr 16 '13 at 08:33 — AlexM, Apr 16 '13 at 08:33

score 5 · Answer 1 · answered Apr 16 '13 at 08:36

You would need to apply De l'Hôpital's Rule six (!) times before obtaining an answer.

From the Taylor expansions, we know $\sin(t^2) = t^2 - \dfrac{t^6}{3!} + O(t^{10})$ and $\exp(t^3) = 1 + t^3 + \dfrac{t^6}{2!} + O(t^9)$.

Thus, the quotient becomes:

$$\lim_{t \to 0} \frac{e^{t^3}-1-t^3}{\sin(t^2)-t^2} = \lim_{t\to 0} \frac{t^6/2! + O(t^9)}{t^6/3! + O(t^{10})} = \frac{3!}{2!} = 3$$

score 4 · Answer 2 · answered Apr 16 '13 at 08:35

4

Hint:
Use Taylor's formula for $x\to{0}$: $$e^{x}=1+x+\dfrac{x^2}{2!}+\ldots+\dfrac{x^n}{n!}+o(x^n), \\ \sin{x}=x-\dfrac{x^3}{3!}+\ldots+(-1)^n\dfrac{x^{2n+1}}{{(2n+1)}!}+o(x^{2n+1}).$$

answered Apr 16 '13 at 08:35

M. Strochyk

8,397

score 2 · Answer 3 · answered Apr 16 '13 at 08:59

Write it up as $$\frac{t^6}{\sin (t^2) - t^2} \cdot \frac{e^{t^3} - 1 - t^3}{t^6 }$$

For first use L'hopital rule three times and you will get this value. Similarly use L'hopital rule 4 times you will get this for the second. Of course, Taylor expansion is much easier.

Tricky Limit question

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