## anonymous 4 years ago 4th question: Evaluate the lim x^2-16 over x+4 as x->4.

|dw:1328533359544:dw|

2. ash2326

we have $lim_{x->4} \frac{x^2-16}{x+4}$ a^2-b^2=(a+b)(a-b) so we get $lim_{x->4} \frac{(x+4)(x-4)}{x+4}$ we get $lim_{x->4} (x-4)$ $4-4=0$

3. anonymous

$\lim_{x->4} \frac{x^2-16}{x+4} = \lim_{x->4} \frac{(x+4)(x-4)}{(x+4)}$$=\lim_{x->4} \frac{\cancel{(x+4)}(x-4)}{\cancel{(x+4)}}=\lim_{x\to 4} x-4=0$

4. anonymous

thanks guys i'll post the other problems