Prove that the following function is continuous at $0$

Question

Prove that the following function is continuous at $0$
$$ f(x) = \begin{cases} \frac{\sin(3x)}{\tan(2x)} \qquad \text{if} \ x<0 \ ; \\ \\ \frac{3}{2} \qquad\quad\qquad \text{if} \ x=0 \ ; \\ \\ \frac{\log(1+3x)}{e^{2x}-1} \ \ \ \text{if} \ x>0 \ . \end{cases} $$

How do I solve this problem?

$\displaystyle\lim_{x\to 0} \frac{\log(1+3x)}{3x} = 1 $ and $\displaystyle\frac{e^{2x}-1}{2x}=\log_ee$ .

Hope this will help.

This isnt my homework. I am just asking my doubts. This is my summer vacation — chndn, Apr 19 '13 at 03:43

vadim123 · Accepted Answer · 2013-04-19T14:11:51.497

2

Some hints (really the same hint twice):

$\displaystyle \frac{\sin(3x)}{\tan(2x)}=\frac{\sin(3x)}{x}\frac{x}{\tan(2x)}$

$\displaystyle \frac{\log(1+3x)}{e^{2x}-1}=\frac{\log(1+3x)}{x}\frac{x}{3^{2x}-1}$

Extra bonus hint:

$\underset{x\rightarrow 0}{\lim}\frac{\sin(x)}{x}=1$

edited Apr 19 '13 at 14:11

answered Apr 19 '13 at 03:44

vadim123

82,796

I got the last one but not the first one. Can u elaborate +1 – chndn Apr 19 '13 at 03:50

score 0 · Answer 2 · answered Apr 19 '13 at 03:23

0

It is not. $\tan(-2 \pi)=0$ so $f(x)$ is not even defined at $x=-\pi$

answered Apr 19 '13 at 03:23

Ross Millikan

374,822

Sorry , I forgot to mention "at 0" – chndn Apr 19 '13 at 03:30
1

@chndn: so take the limit of the values for $x$ approaching $0$ from below and show they approach $\frac 32$ and again from above. Where are you having trouble? You could look at the Taylor series. Your statement that $\displaystyle\lim_{x\to 0} \frac{\log(1+3x)}{1+3x} = 1$ is not correct. The numerator goes to zero and the denominator goes to 1, so the limit is $0$ – Ross Millikan Apr 19 '13 at 03:33
How do I get $\lim_{x\to 0}\frac{sin3x}{tan2x} = \frac{3}{2}$? and question edited. thanks...+1 – chndn Apr 19 '13 at 03:36
@chndn: I use the small angle approximation $\sin x = \tan x =x$ for $x \ll 1$ but I am a physicist/engineer. I don't know if that is OK for you. – Ross Millikan Apr 19 '13 at 03:39
I didnt quite understand your observation. I want to solve it in mathematical steps( or something like that) – chndn Apr 19 '13 at 03:41
You have changed the limit you cite. The earlier one does what I said before. Now your statement is correct. user73324 has given some more explicit hints. Can you use them? – Ross Millikan Apr 19 '13 at 03:49
yah , i got the second one. I couldnt get the first – chndn Apr 19 '13 at 03:51
Do you know $\lim_{x \to 0} \frac {\sin x}x=1$? Then use L'Hopital or the chain rule or multiple angle identities. – Ross Millikan Apr 19 '13 at 03:55
I would be curious about the reason for the downvote. When I posted it just asked about why $f(x)$ was not continuous-it didn't specify "at $0$". My answer shows why it is not. – Ross Millikan Apr 19 '13 at 03:57
how do I solve $\frac{3x}{tan2x}$ and I had upvoted – chndn Apr 19 '13 at 03:58
@chndn: As I said before, you can use multiple angle identities if you don't have other tools. Expand $\tan 2x$ and recognize that $\cos x$ goes to $1$ as $x \to 0$. – Ross Millikan Apr 19 '13 at 04:01
what is the identity – chndn Apr 19 '13 at 04:01
1

The usual one is $\tan 2x=\frac {2 \tan x}{1-\tan^2 }$ – Ross Millikan Apr 19 '13 at 04:05
How can I use it here? can you solve it for me? – chndn Apr 19 '13 at 04:06
Please give it some thought. – Ross Millikan Apr 19 '13 at 04:07
I didnt get it. whatever I do i cant make it 3/2 – chndn Apr 19 '13 at 04:08
$\sin 3x \approx 3x, \tan 2x =\cos 2x \sin 2x \approx 1 \cdot 2x$ – Ross Millikan Apr 19 '13 at 04:12
okay got it thanks – chndn Apr 19 '13 at 11:23

Prove that the following function is continuous at $0$

2 Answers2