Find limit of function as $x$ approaches $ 0$ Find limit of $$\mathop{\lim}\limits_{x \to 0}6x^2\times(\cot(x))(\csc(2x))$$
Can somebody please explain it all throughly. For some reason this seems to be very confusing for me.
 A: $$6x^2 \cot x \csc (2x)$$
$$6x^2 \frac{\cos x}{\sin x \sin (2x)}$$
$$6x^2 \frac{\cos x}{\sin x (2\sin x \cos x)}$$
$$\frac{6x^2}{2\sin^2 x}$$
$$\frac{3}{\frac{\sin^2 x}{x^2}}$$
We know that, Limit of $\frac{\sin x}{x} = 1$, when $x$ tends to $0$.
Hence the value of given limit is: $3$
Note: If you do not know that Limit of $\frac{\sin x}{x} = 1$, when $x$ tends to $0$ and you wish to prove it, you can either prove it by using L'Hospital rule or by using Taylor series expansion for $\sin x$.
A: $\csc(2x)=1/\sin(2x)$
So we have,
$$6x^2\cdot (\cos(x)/\sin(x))\cdot 1/\sin(2x)$$
Or,
$(6x^2\cdot \cos(x))/(\sin(x)\sin(2x)$
Because after putting $x=0$, we have $0/0$,we have to use L'Hôpital's rule, or make derivative up part and below part, and then trying to put $x=0$ again, could you continue from this?
http://mathworld.wolfram.com/LHospitalsRule.html
A: First of all, the product of the trigonometric function simplifies to $\dfrac{\csc ^2(x)}{2}$. Then, truncated to the first terms, the Taylor series of $\csc(x)$ is $\frac{1}{x}+\frac{x}{6}+O\left(x^3\right)$. So, replace and expand the expression : you should arrive to $$3+x^2+\frac{x^4}{5}$$ So the limit is ???
