# Limits next question

1. Jul 6, 2009

### Laven

$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}$$

so this is the question.

I'm here solving this problem you please check where am i wrong or next idea i've to use here.
=$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}x\frac{cos{x}+cos{y}}{cos{x}+cos{y}}$$
=$$\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{{x-y}{cos{x}+cos{y}}$$
=$$\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{{x-y}{cos{x}+cos{y}}$$ +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}[/tex]

after this i don't 've idea wht to do.Is there next idea we have to include overhere?

2. Jul 6, 2009

### HallsofIvy

Staff Emeritus
The "x" you had in the middle here was just "times" wasn't it? Better not to use such a symbol along with"x" as a variable.

$$=[tex]\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}$$
=$$\lim_{x\rightarrow y}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\frac{sin(x-y)}{(x-y)(cos{x}+cos{y})}$$

after this i don't 've idea wht to do.Is there next idea we have to include overhere?[/QUOTE]
I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
$$\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}$$
which goes to 1 as x goes to y, and
$$\frac{sin x cos x- sin y cos y}{x- y}$$
which is still a problem.

It would be far simpler to use L'Hopital's rule. Are you allowed to do that?

3. Jul 7, 2009

### Laven

$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}$$

so this is the question.

I'm here solving this problem you please check where am i wrong or next idea i've to use here.
=$$\lim_{x\rightarrowy}\frac{sin{x}-sin{y}}{x-y}*\frac{cos{x}+cos{y}}{cos{x}+cos{y}}$$
=$$\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x}cos{y}-cos{x}sin{y}}{(x-y)(cos{x}+cos{y})}$$
=$$\lim_{x\rightarrowy}\frac{sin{x}cos{x}-sin{y}cos{y}+sin{x-y}}{(x-y)(cos{x}+cos{y})} +\fracsin{x-y}{(x-y)(cos{x}+cos{y})}$$

after this i don't 've idea wht to do.Is there next idea we have to include over
here?

4. Jul 7, 2009

### Laven

I think I have corrected your LaTex properly. sin(x- y)= sin(x)cos(y)- cos(x)sin(y) and you appear to be trying to put the numerator in that form. (It would be a good idea to explain things like that when asking about a problem.) You wind up with two parts:
$$\frac{sin x cos y- cos x sin y}{x- y}= \frac{sin(x-y)}{x- y}$$
which goes to 1 as x goes to y, and
$$\frac{sin x cos x- sin y cos y}{x- y}$$
which is still a problem.

It would be far simpler to use L'Hopital's rule. Are you allowed to do that?[/QUOTE]

Yea that's what i meant to express in latex form.Thanks for your correction.

Isn't it solved by that method?[method which i did]

Umm..L'Hopital's rule is it the best way to solve the problem?Thanks I don't 've idea bout that but i can get it after reading this rule.If not i'll again ask you.

5. Jul 9, 2009

### tiny-tim

Hi Laven!

You need to learn your trigonometric identities …

in this case sinx - siny = 2 sin((x - y)/2) cos((x + y)/2)