# Thread: limits with trig function test prep

1. ## limits with trig function test prep

I am preparing for a test and the following question is from the practice exam. My problem is that the explanation skips over things that I don't understand. I was hoping somebody here could clear it up for me.

Find the limit:
lim x(sin 5x) / (1 − cos x) = ?
x→0

I know that : x sin x / x = 1
and : 1-cos x / x^2 = 1/2

And these come into play in this problem.. .it's the "getting there" I don't understand. here's the solution given:

lim x(sin 5x) / (1 − cos x) = lim 5x^2(sin 5x) / 5x(1-cosx) = 5/(1/2)
x→0 x→0

The solution never shows where the known things above actually appear. where is the x that x sin x is divided by? Where is the x^2 that 1-cosx is divided by?

2. Hello, littlejodo!

Find: .$\displaystyle \lim_{x\to0}\frac{x\sin5x}{1-\cos x}$

I know that: .$\displaystyle \lim_{x\to0}\frac{\sin x}{x}\:=\:1\;\text{ and }\;\lim_{x\to0}\frac{1-\cos x}{x^2} \:=\:\tfrac{1}{2}$ . . . . Good!

We have: .$\displaystyle \frac{x\sin5x}{1-\cos x}$

Divide top and bottom by $\displaystyle x^2\!:\quad\frac{\dfrac{x\sin5x}{x^2} }{\dfrac{1-\cos x}{x^2}} \;=\;\frac{\dfrac{\sin 5x}{x}}{\dfrac{1-\cos x}{x^2}}$

Multiply numerator by $\displaystyle \frac{5}{5}\!:\quad \frac{\frac{5}{5}\cdot\dfrac{\sin5x}{x}}{\dfrac{1-\cos x}{x^2}} \;=\;\frac{5\cdot\dfrac{\sin5x}{5x}}{\dfrac{1-\cos x}{x^2}}$

Therefore: .$\displaystyle \lim_{x\to0}\left[\frac{5\cdot\dfrac{\sin5x}{5x}} {\dfrac{1-\cos x}{x^2}}\right] \;=\; \frac{5\cdot1}{\frac{1}{2}} \;=\;10$

3. thank you so much!!