## anonymous 5 years ago Find the lim x->-infinity of (6-7x)/(3+2x)^4

1. watchmath

It is of the type $$\frac{\infty}{\infty}$$ so we can use L'Hospital rule and reduce the limit into: $$\lim_{x\to -\infty}\frac{-7}{4(3+2x)^3}=0$$

2. watchmath

sorry wrong derivative on the bottom :)

3. anonymous

yeah i saw that but either way it doesnt matter.

4. watchmath

the bottom should be $$8(3+2x)^3$$, but the limit is still 0.

5. myininaya

$\lim_{x \rightarrow -\inf}\frac{6-7x}{(3+2x)^4}\times \frac{x^4}{x^4}$

6. myininaya

top goes to 0 bottom gores to 2^4 so the limit is 0

7. myininaya

i meant to write (1/x^4)/(1/x^4)

8. anonymous

The fraction with the denominator expanded. $\frac{6-7x}{\left(81+216 x+216 x^2+96 x^3+16 x^4\right)}$ It appears that the 16x^4 term along will do the job.

9. myininaya

yeah rob i didn't feel like expanding it lol

10. anonymous

along should have been spelled alone. Sorry.