Question

lim [(sin(x^2+2)-sin(x+2))/x] when x goes to 0. I should use sin(a+b) = sinacosb + sinbcosa?

Answer 1

Are you allowed to use L'hopital's rule?

Answer 2

\[\Large \lim_{x \rightarrow 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}\]

Answer 3

If you know L'hopital's rule, this problem is pretty simple, but there's a possibility you haven't been taught that yet. Let me know if you're allowed.

Answer 4

Yeah, I'm allowed to use L'Hôspital rule, but doing in "pure algebra" is nice to learn, I think. ^^

Answer 5

How do you wrote the lim in a bigger font

Answer 6

?

Answer 7

Use the equation editor.
and I typed:
Large lim_{x rightarrow 0} frac{sin(x^2+2)-sin(x+2)}{x}

Large makes it bigger, and typing Large makes it larger than typing large.
The only other part that might be hard to figure out is:
frac{numerator}{denominator} makes a fraction

Answer 8

Thanks!

Answer 9

You're welcome. Okay, so here's how to use L'hopital's on this.
First, make sure that you're allowed to use the rule. Is the function of indeterminate form? Well, if we put in x=0, we get lim = 0/0, so yes, we can use L'hopital's rule.
Simply take the derivative of top and bottom:
\[\Large = \lim_{x \rightarrow 0} \frac{2x*\cos(x^2+2)-\cos(x+2)}{1} = -\cos(2)\]

Answer 10

Now, as for solving it algebraically, I think you're on the right track... but that's a good amount of work, and frankly, I don't want to do it, since I can just L'hopital this bad boy right in the throat.

Answer 11

Ah it's true that is much easier use L'Hôspital in this case. Thanks!!
A doubt : I'm brazilian, and I don't understand very nice this: "right in the throat." is this only a expression that means accept something? I'm curious.

Answer 12

lol sorry about that. That is not a common phrase at all, and hardly worth explaining.

Answer 13

I just meant to express that it is kind of an easy solution to a difficult problem.

Answer 14

Ahh right ! Rsrs thanks again!