MHB How Do You Calculate the Limit of 3sin(5x)/sin(3x) as x Approaches 0?

[Aychetown]

I was on khan academy and I'm beginning to relearn calculus for the first time in years and I couldn't figure out how to prove this function. I think I was supposed to use my calculator but I would like to know if anyone can work it out for me?

It was the limit as x-0 of 3sin(5x)/sin(3x). The answer is supposed to be 7.5 but I can't figure out how to work it out. Help me my fellow math heads.

 

[Opalg]

I was on khan academy and I'm beginning to relearn calculus for the first time in years and I couldn't figure out how to prove this function. I think I was supposed to use my calculator but I would like to know if anyone can work it out for me?

It was the limit as x-0 of 3sin(5x)/sin(3x). The answer is supposed to be 7.5 but I can't figure out how to work it out. Help me my fellow math heads.

Hi Aychetown and welcome to MHB!

In this problem, you are probably meant to start from the fact that $$\lim_{x\to0}\frac{\sin x}x = 1.$$ If you replace $x$ by $5x$, you get $$\lim_{x\to0}\frac{\sin (5x)}{5x} = 1.$$ And if you then multiply that by $5$, it becomes $$\lim_{x\to0}\frac{\sin (5x)}{x} = 5$$ (and the same thing would work if you replace the $5$ by any other multiple).

Now suppose you write your limit as $$\lim_{x\to0}3\frac{\sin(5x)/x}{\sin(3x)/x}$$ (dividing top and bottom by $x$). You can then take the limits of the numerator and denominator separately, to get the answer. But the answer will not be 7.5, because you seem to have stated the problem incorrectly. My guess is that it should be $$\lim_{x\to0}3\frac{\sin(5x)}{\sin(2x)}$$, with a $2$ in the denominator.