# LIMxgoesto0(tan(7x)cosec(3x))

• April 15th 2010, 01:35 AM
Nicholas
LIMxgoesto0(tan(7x)cosec(3x))
any plan of attack would be great. :)
• April 15th 2010, 01:55 AM
Failure
Quote:

Originally Posted by Nicholas
any plan of attack would be great. :)

If you know that $\sin(ax)=ax+o(x)$ and $\cos(ax)= 1-ax+o(x)$ for $x\to 0$, you are almost done:

$\lim_{x\to 0}\left(\tan(7x)\cdot \mathrm{cosec}(3x)\right)=\lim_{x\to0}\frac{\sin(7 x)}{\cos(7x)\cdot \sin(3x)}$

$=\lim_{x\to 0}\frac{7x+o(x)}{(1-7x+o(x))\cdot (3x+o(x))}=\lim_{x\to 0}\frac{7x}{(1-7x)\cdot 3x}=\ldots$

Where $o(x)$ is Landau's little-oh notation: $\lim_{x\to 0}\frac{o(x)}{x}=0$.
• April 15th 2010, 01:55 AM
Soroban
Hello, Nicholas!

You're expected to know this theorem:

. . $\lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;\lim_{\theta\to0}\frac{\theta}{\sin\theta} \:=\:1$

Quote:

$\lim_{x\to0}\tan7x\csc3x$

We have: . $\tan7x\csc3x \:=\:\frac{\sin7x}{\cos7x}\cdot\frac{1}{\sin3x}$

Multiply by $\frac{7x}{7x}$ and $\frac{3x}{3x}\!:\quad {\color{blue}\frac{7x}{7x}}\cdot\frac{\sin7x}{\cos 7x} \cdot {\color{blue}\frac{3x}{3x}}\cdot\frac{1}{\sin3x}$

. . $=\;\frac{7x}{1}\cdot\frac{\sin7x}{7x}\cdot\frac{1} {\cos7x} \cdot \frac{1}{3x}\cdot\frac{3x}{\sin3x} \;=\;\frac{7x}{3x}\cdot\frac{\sin7x}{7x}\cdot\frac {1}{\cos7x} \cdot \frac{3x}{\sin3x}$

Therefore: . $\lim_{x\to0}\left(\frac{7}{3}\cdot\frac{\sin7x}{7x }\cdot\frac{1}{\cos7x}\cdot\frac{3x}{\sin3x}\right ) \;=\;\frac{7}{3}\cdot 1\cdot\frac{1}{1}\cdot 1 \;=\;\frac{7}{3}$

• April 15th 2010, 02:00 AM
Nicholas
Apologies, I am a first year uni student and have not yet learnt that theorm. Thankyou very much though for explaining it! Much appreciated. :)