# One sided limit question

• Nov 6th 2010, 01:42 PM
Vamz
One sided limit question
$\displaystyle \frac{12sin^3x}{12x^3}$
$\displaystyle \lim x \to 0^-$

Heres what I have...
can be rewritten as:
$\displaystyle \frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1$

Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a peicewise function?

Thanks!
• Nov 6th 2010, 02:13 PM
mr fantastic
Quote:

Originally Posted by Vamz
$\displaystyle \frac{12sin^3x}{12x^3}$
$\displaystyle \lim x \to 0^-$

Heres what I have...
can be rewritten as:
$\displaystyle \frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1$

Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a peicewise function?

Thanks!

$\displaystyle \displaystyle \lim_{x \to 0^-} \frac{12 \sin^3 (x)}{12 x^3} = \lim_{x \to 0^+} \frac{12 \sin^3 (x)}{12 x^3} = \lim_{x \to 0} \frac{12 \sin^3 (x)}{12 x^3} = 1$.
• Nov 6th 2010, 02:19 PM
Quote:

Originally Posted by Vamz
$\displaystyle \frac{12sin^3x}{12x^3}$

$\displaystyle \lim x \to 0^-$

Heres what I have...
can be rewritten as:

$\displaystyle \frac{12(sinx)^3}{12x^3}=\frac{(sinx)^3}{x^3}=\fra c{sinx}{x}*\frac{sinx}{x}*\frac{sinx}{x}=1*1*1=1$

Up to this point, am I correct? How does the fact that it wants the limit as x approaches zero from the left? I only know how to work with these one-sided limits in a peicewise function. What do I do from here & what do I need to learn to solve one sided limits that are NOT in a piecewise function?

Thanks!

$\displaystyle x=-u$

$\displaystyle x\rightarrow\ 0^-\Rightarrow\ u\rightarrow\ 0^+$

$\displaystyle Sinx=Sin(-u)=-Sinu$

$\displaystyle \displaystyle\frac{Sinx}{x}=\frac{-Sinu}{-u}=\frac{Sinu}{u}$