Question

Find the lim x->-infinity of (6-7x)/(3+2x)^4

Answer 1

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

Answer 2

sorry wrong derivative on the bottom :)

Answer 3

yeah i saw that but either way it doesnt matter.

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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.

Answer 9

yeah rob i didn't feel like expanding it lol

Answer 10

along should have been spelled alone. Sorry.