# Math Help - Limits with trig values

1. ## Limits with trig values

I'm having trouble with my AP Calculus homework dealing with trig limits. I get these long values and have no idea how to reduce them.

Example:

Limit as x approaches zero of cot(3x)/csc(2x)

I understand the first part:

Lim x->0 (cos(3x)/sin(3x))/(1/sinx)

which reduces to:

Lim x -> 0 (cos(3x)sin(x))/sin(3x)

but I'm at a loss of where to go from here. Everything I try either becomes so complicated I don't know what to do with it or ends up as 0/0. Thanks in advance to anyone who can shed some light on this dilemma.

2. Originally Posted by oObutterfly-chaserOo
I'm having trouble with my AP Calculus homework dealing with trig limits. I get these long values and have no idea how to reduce them.

Example:

Limit as x approaches zero of cot(3x)/csc(2x)

I understand the first part:

Lim x->0 (cos(3x)/sin(3x))/(1/sinx)

which reduces to:

Lim x -> 0 (cos(3x)sin(x))/sin(3x)

but I'm at a loss of where to go from here. Everything I try either becomes so complicated I don't know what to do with it or ends up as 0/0. Thanks in advance to anyone who can shed some light on this dilemma.
Hopefully you know that $\lim_{x \to 0}\frac{\sin(ax)}{ax}=1$ for all $a \ne 0$

using this fact and some trig identities

$\frac{\cot(3x)}{\csc(2x)}=\frac{\cos(3x)}{\sin(3x) }\cdot \sin(2x)$

Now multiply both the numerator and denominator by 6x to get

$\frac{6x}{6x}\frac{\cos(3x)}{\sin(3x)}\cdot \sin(2x)=\frac{2}{3}\cos(3x) \cdot \frac{3x}{\sin(3x)}\cdot \frac{\sin(2x)}{2x}$

using the above we get

$\lim_{x\to 0}\frac{2}{3}\cos(3x) \cdot \frac{3x}{\sin(3x)}\cdot \frac{\sin(2x)}{2x}=\frac{2}{3}\cdot 1 \cdot 1 \cdot 1=\frac{2}{3}$

3. Originally Posted by oObutterfly-chaserOo
I'm having trouble with my AP Calculus homework dealing with trig limits. I get these long values and have no idea how to reduce them.

Example:

Limit as x approaches zero of cot(3x)/csc(2x)

I understand the first part:

Lim x->0 (cos(3x)/sin(3x))/(1/sinx)

which reduces to:

Lim x -> 0 (cos(3x)sin(x))/sin(3x)

but I'm at a loss of where to go from here. Everything I try either becomes so complicated I don't know what to do with it or ends up as 0/0. Thanks in advance to anyone who can shed some light on this dilemma.
L'Hospital's Rule works quite nicely here, since your limit is of the indeterminate form $\frac{0}{0}$.

$\lim_{x\to 0}\frac{\cos{(3x)}\sin{(2x)}}{\sin{(3x)}}$

$= \lim_{x \to 0}\frac{2\cos{(2x)}\cos{(3x)} - 3\sin{(2x)}\sin{(3x)}}{3\cos{(3x)}}$

$= \frac{2}{3}$.