Limits Urgent!

**iz1hp** · September 27th 2008, 12:15 PM

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

**Jhevon** · September 27th 2008, 12:22 PM

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")

**iz1hp** · September 27th 2008, 12:29 PM

what do you mean special limit?

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

**Jhevon** · September 27th 2008, 12:32 PM

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 $\lim_{x \to 0^-} \frac {\sin x}x = 1$

**iz1hp** · September 27th 2008, 12:35 PM

yes i know that.

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

**Jhevon** · September 27th 2008, 12:37 PM

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

**iz1hp** · September 27th 2008, 12:45 PM

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 $

and the answer is 0?

**bkarpuz** · September 27th 2008, 12:49 PM

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 $

and the answer is 0?

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.

**Jhevon** · September 27th 2008, 12:49 PM

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 $

and the answer is 0?

yes, the answer is zero, but i do not like how you phrased it. what do you mean "cancels"? we didn't cancel anything here. and where is your limit? you have to take the limit to see what happens. you can't just equate things like that and say this is equal to that. it doesn't work

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