Trouble solving this limit without using l'Hôpital's rule

Question

The limit I want to calculate is the following $$ \lim_{x \to 0}{\frac{(e^{\sin(4x)}-1)}{\ln\big(1+\tan(2x)\big)}} $$ I've been stuck on this limit for a while and I don't know how to solve it please help me.

You'd have more trouble doing this with the Hospital's rule. The differentiations are awfully tedious. — Angina Seng, Nov 10 '18 at 16:48
One general strategy is to write out the first few term of the Taylor series expansions of the ingredients and do the algebra to find the first few terms of the series for the full expression. (Not posted as an answer because I haven't tried it.) — Ethan Bolker, Nov 10 '18 at 16:51

S. Maths · Answer 1 · 2020-02-04T01:08:42.957

10

Hint: $$\frac{(e^{\sin 4x}-1)}{\ln(1+\tan 2x)}= \frac{(e^{\sin 4x}-1)}{\sin 4x}\frac{\tan 2x}{\ln(1+ \tan 2x)} \frac{\sin 4x}{4x} \frac{2x}{\tan 2x} \times 2.$$

edited Feb 04 '20 at 01:08

answered Nov 10 '18 at 16:48

S. Maths

2,238

hey thank you for the hint , so does this limit = 2? – Levio Nov 10 '18 at 16:58
@Levio, Yeah, your right. – S. Maths Nov 10 '18 at 17:00
The omitted signs between fraction confused me for a second. I'd better get some sleep. – Oleg Lobachev Nov 10 '18 at 21:52

score 3 · Answer 2 · edited Nov 10 '18 at 19:43

3

As $x\to 0$, $$\exp(\sin 4x)-1\sim\sin4x\sim 4x$$ and $$\ln(1+\tan 2x)\sim\tan2x\sim 2x$$ etc.

edited Nov 10 '18 at 19:43

John Coleman

5,401

answered Nov 10 '18 at 16:50

Angina Seng

158,341

score 2 · Answer 3 · answered Nov 10 '18 at 16:55

2

Hint: Make use the following facts: 1. $\dfrac{e^{\sin(4x)} - 1}{\sin(4x)} \to 1$

2.$\dfrac{\log(1 + \tan(2x))}{\tan(2x)} \to 1$

$\cos(2x) \to 1$

answered Nov 10 '18 at 16:55

DeepSea

77,651

thanks for the hints I managed to solve it – Levio Nov 10 '18 at 17:01

Trouble solving this limit without using l'Hôpital's rule

3 Answers3