Gosh, this is so blasted hard to read! [b]Everyone[/b], if you care at all about people here helping you,[b]type[/b] the problems in, do not post unreadable photos of the problem!

I [i]think[/i] that the problem you are asking about is to find [tex]\lim_{x\to 0}\frac{sin(5x)}{2x(x- 2)}[/tex]. And you did this by writing it as [tex]\frac{1}{2}\lim_{x\to 0}\frac{sin(5x)}{x}\frac{1}{x-2}= \frac{5}{2}\left(\lim_{x\to 0}\frac{sin(5x)}{5x}\right)\left(\lim_{x\to 0}\frac{1}{x- 2}\right)[/tex].

You [b]can[/b] do those two limits separately because "[tex]\lim_{x\to a}f(x)g(x)= \left(\lim_{x\to a}f(x)\right)\left(\lim_{x\to a} g(x)\right)[/tex]" Of course, [tex]\lim_{x\to 0}\frac{1}{x- 2}= \frac{1}{0- 2}= -\frac{1}{2}[tex] and [tex]\lim_{x\to 0}\frac{sin(5x)}{5x}= 1[/tex] so the original limit is (5/2)(-1/2)(1)= -5/4.

The image is very unclear.
Could you please write which limit you would like to be solved?

My teacher said I have to solve all the limits at once, but I don't understand why it is wrong to solve one limit before the other.
[attachment=0]test picture.jpg[/attachment]