Trigonometric limit

[Imum Coeli]

I need to find the limit as x goes to zero of sin(3x)/sin(5x).

I have a hint to write sin(3x)/sin(5x) in the form A(sin(y)/y)/(sin(z)/z) but I really have no idea what to do. I guess I'm supposed to use a trig identity but I can't figure out which one.

Thanks.

 

[Deleted member 4993]

I need to find the limit as x goes to zero of sin(3x)/sin(5x).

I have a hint to write sin(3x)/sin(5x) in the form A(sin(y)/y)/(sin(z)/z) but I really have no idea what to do. I guess I'm supposed to use a trig identity but I can't figure out which one.

Thanks.

In addition to the hint - use the fact that:

\(\displaystyle \displaystyle \lim_{y \to 0}\frac{sin(y)}{y} \ = \ 1\)

 

[Imum Coeli]

In addition to the hint - use the fact that:

\(\displaystyle \displaystyle \lim_{y \to 0}\frac{sin(y)}{y} \ = \ 1\)

I know that.

The concepts of limits themselves and a few methods (The squeeze theorem and L'hopital's rule) don't present too much difficulty for me. I just have a real problem with writing "The same thing in different words"...

This problem is driving me nuts because I'm sure it is quite easy. I'm just not seeing it.

 

[Imum Coeli]

Using L'Hopitals rule I get an answer of 3/5. But at no point have I written anything in the form A(sin(y)/y)/(sin(z)/z). How do I get it to this form?

 

[Deleted member 4993]

Using L'Hopitals rule I get an answer of 3/5. But at no point have I written anything in the form A(sin(y)/y)/(sin(z)/z). How do I get it to this form?

\(\displaystyle \dfrac{sin(3x)}{sin(5x)}\)

\(\displaystyle \dfrac{sin(3x)}{3x} \ * \ \dfrac{5x}{sin(5x)} \ * \ \dfrac{3x}{5x}\)

Now what ....

 

[soroban]

Hello, Imum Coeli!

We are expected to know that: .\(\displaystyle \displaystyle\lim_{\theta\to0}\frac{\sin \theta}{\theta} \:=\:1\)

\(\displaystyle \displaystyle\lim_{x\to0}\frac{\sin(3x)}{\sin(5x)}\)

This is what Subhotosh's advice means . . .

We have: .\(\displaystyle \displaystyle\frac{\sin(3x)}{\sin(5x)}\)

Divide numerator and denominator by \(\displaystyle \displaystyle x\!:\; \frac{\frac{\sin(3x)}{x}}{\frac{\sin(5x)}{x}}\)

Multiply the numerator by \(\displaystyle \frac{3}{3}\), the denominator by \(\displaystyle \frac{5}{5}\!:\)
. . \(\displaystyle \displaystyle \frac{\frac{3}{3}\cdot\frac{\sin(3x)}{x}}{\frac{5}{5}\cdot\frac{\sin(5x)}{x}} \;=\;\frac{3\cdot\frac{\sin(3x)}{3x}}{5\cdot\frac{\sin(5x)}{5x}} \;=\;\frac{3}{5}\cdot\frac{\frac{\sin(3x)}{3x}}{ \frac{\sin(5x)}{5x}} \)

Therefore: .\(\displaystyle \displaystyle \lim_{x\to0}\left[\frac{3}{5}\cdot\frac{\frac{\sin(3x)}{3x}}{ \frac{\sin(5x)}{5x}}\right] \;=\;\frac{3}{5}\cdot\frac{1}{1} \;=\;\frac{3}{5} \)

 

[Imum Coeli]

Aha! I see now. Thank you both so much!