# Math Help - Limits Urgent!

1. ## Limits Urgent!

lim x->0 from the left of (2x + 3 + (sin3x/|x|))

2. Originally Posted by iz1hp
lim x->0 from the left of (2x + 3 + (sin3x/|x|))
Note that as $x \to 0^-$ we have $x < 0$ so that $|x| = -x$. so you want

$\lim_{x \to 0^-} \bigg( 2x + 3 + \frac {\sin 3x}{-x} \bigg)$

(think "special limit")

3. what do you mean special limit?

i got to changing the |x| to -x but i dont know where to go from there....

4. Originally Posted by iz1hp
what do you mean special limit?

i got to changing the |x| to -x but i dont know where to go from there....
there is a special limit that says $\lim_{x \to 0} \frac {\sin x}x = 1$. since the limit exists, then of course the one sided limits exist as well, so that we know $\lim_{x \to 0^-} \frac {\sin x}x = 1$

5. yes i know that.

but does that mean i have to multiply all the values within this equation by 3 over 3?

6. Originally Posted by iz1hp
yes i know that.

but does that mean i have to multiply all the values within this equation by 3 over 3?
not all, just the one with sine. why bother the others? we have no problem with them

7. so you multiply by 3 over 3 and the equation becomes

$
\bigg( 2x + 3 + \frac {3 sin 3x}{-3x} \bigg)
$

and then it cancels and you get

$
2x + 3 - 3
$

8. Originally Posted by iz1hp
so you multiply by 3 over 3 and the equation becomes

$
\bigg( 2x + 3 + \frac {3 sin 3x}{-3x} \bigg)
$

and then it cancels and you get

$
2x + 3 - 3
$

Note that $-3\frac{\sin(3x)}{3x}\neq-3$.
$-3\frac{\sin(3x)}{3x}\to-3$ is true when $x\to0$.
So you can not cancel it.
Also see the post below posted a sec after me.

9. Originally Posted by iz1hp
so you multiply by 3 over 3 and the equation becomes

$
\bigg( 2x + 3 + \frac {3 sin 3x}{-3x} \bigg)
$

and then it cancels and you get

$
2x + 3 - 3
$