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

any plan of attack would be great.

2. Originally Posted by Nicholas
any plan of attack would be great.
If you know that $\displaystyle \sin(ax)=ax+o(x)$ and $\displaystyle \cos(ax)= 1-ax+o(x)$ for $\displaystyle x\to 0$, you are almost done:

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

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

Where $\displaystyle o(x)$ is Landau's little-oh notation: $\displaystyle \lim_{x\to 0}\frac{o(x)}{x}=0$.

3. Hello, Nicholas!

You're expected to know this theorem:

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

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

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

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

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

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

4. Apologies, I am a first year uni student and have not yet learnt that theorm. Thankyou very much though for explaining it! Much appreciated.

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### lim tan7x cosec3x

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