Limits of Trig Ratios Questions

[CreativeUsername]

Specifically:

1/cot^2(x)/x as x approaches 0

All I've been able to do so far has been along the lines of:
1/(cos^2/sin^2)/x
(tan^2)(1/x)
(sin^2/cos^2)(1/x)
(sin^2/x*cos^2)
sin^2/x*(1-sin^2)
sin^2/x-xsin^2

Which is just 0/(0-0(0)). I know the answer is 1, but I haven't been able to get anything that works out to that. Any help would be appreciated.

I was also bothered by sin(2x)/sin(3x). The worksheet it was from says it can be written as (2 sin(2x)/2x)*(3x/3sin(3x)), but I can't tell why and my teacher never went over the sheet.

 

[soroban]

Hello, CreativeUsername!

Specifically: .1/cot^2(x)/x as x approaches 0.

Is that: .\(\displaystyle \dfrac{\:\left(\dfrac{1}{\cot^2x}\right)\:}{x}\: \text{ or }\:\dfrac{\quad1\quad}{\left(\dfrac{\cot^2x}{x} \right)} \)

\(\displaystyle \displaystyle\lim_{x\to0}\dfrac{\sin2x}{\sin3x}\)

We will use this theorem: .\(\displaystyle \displaystyle \lim_{\theta\to0}\frac{\sin\theta}{\theta} \;=\;\lim_{x\to0}\frac{\theta}{\sin\theta} \;=\;1\)


We have: .\(\displaystyle \dfrac{\sin2x}{1}\cdot\dfrac{1}{\sin3x}\)

Then: .\(\displaystyle \dfrac{\sin2x}{1}\cdot\dfrac{2x}{2x}\cdot\dfrac{1}{\sin3x}\cdot\dfrac{3x}{3x} \;=\;\dfrac{2x}{3x}\cdot\dfrac{\sin2x}{2x}\cdot \dfrac{3x}{\sin3x}\)


Therefore: .\(\displaystyle \displaystyle \lim_{x\to0}\left[\frac{2}{3}\cdot\frac{\sin2x}{2x}\cdot\frac{3x}{ \sin3x}\right] \;=\;\frac{2}{3}\cdot1\cdot1 \;=\;\frac{2}{3}\)

 

[CreativeUsername]

Hello, CreativeUsername!


Is that: .\(\displaystyle \dfrac{\:\left(\dfrac{1}{\cot^2x}\right)\:}{x}\: \text{ or }\:\dfrac{\quad1\quad}{\left(\dfrac{\cot^2x}{x} \right)} \)

The second one seems more likely, but I'm just guessing since the original didn't have parentheses. Trying both of those on my calculator seems to be getting a limit of 0, though, so it's possible that I might have just misremembered/misheard the answer being one.


Thanks for the help with the other problem. I'd actually never seen the trick of only multiplying one part of a fraction before, but it makes sense.

EDIT: I think I actually figured out the first problem. Does this work? (I cut (x) out of all the trig ratios to make it more legible; they're all still of x)
1/cot²/x
1/(1/tan²)/x
(1/x)(tan²/1)
(tan²/x)
(sin²/cos²)/x
(sin²/x) * (1/cos²)
(1-cos²/x) * (1/cos²)
(((1+cos)*(1-cos))/x) * (1/cos²)

(1-cos)/x = 0

((1+cos)*0) * (1/cos²)
((1+1)0) * (1/1²)
0*1
0

 

[soroban]

Hello, CreativeUsername!

If the first one is: \(\displaystyle \displaystyle\lim_{x\to0}\frac{\:\left(\dfrac{1}{ \cot^2x} \right)\:}{x}\)

We have: \(\displaystyle \displaystyle\frac{1}{x\:\!\cot^2x} \;=\;\frac{\cos^2x}{x\:\!\sin^2x} \;=\;
\frac{x^2}{x^2}\cdot\frac{\cos^2x}{x\;\!\sin^2x} \;=\; \frac{x^2}{\sin^2x}\cdot\frac{\cos^2x}{x^3} \)

Therefore: .\(\displaystyle \displaystyle\lim_{x\to0}\left(\frac{x}{\sin x}\right)^2\cdot\frac{\cos^2x}{x^3} \;=\;1^2\cdot\frac{1^2}{0^3} \;=\;\infty \)



If the first one is: \(\displaystyle \displaystyle\lim_{x\to0}\frac{\quad1\quad}{\left(\frac{\cot^2x}{x} \right)} \)

We have: \(\displaystyle \displaystyle\frac{1}{\frac{\cos^2x}{x\;\!\sin^2x}} \;=\; \frac{x\;\!\sin^2x}{\cos^2x} \;=\; \frac{x^2}{x^2}\cdot\frac{x\;\!\sin^2x}{\cos^2x} \;=\;\frac{\sin^2x}{x^2}\cdot\frac{x^3}{\cos^2x} \)

Therefore: .\(\displaystyle \displaystyle\lim_{x\to0}\left(\frac{\sin x}{x}\right)^2\cdot\frac{x^3}{\cos^2x} \;=\;1^2\cdot\frac{0^3}{1^2} \;=\;0\)