Question

lim 3x^7+x^2
x->-inf

Answer 1

-infinity

Answer 2

So you don't have to do the squeeze theorem or anything

Answer 3

no

Answer 4

Is this true for all polynomials or just this case

Answer 5

with limits to infinity or -infinity?

Answer 6

could explain both cases

Answer 7

for a polynomial (with limits to - or + infinity) just look at what happens to the highest power term.

Answer 8

\[\lim_{x\to-\infty}3x^7=-\infty\]

therefore
\[\lim_{x\to-\infty}3x^7+x^2=-\infty\]

Answer 9

what if the power is negative 3x^-7+x^2 would 2 be the higher power or -7

Answer 10

then it is not a polynomial and

Answer 11

\[\lim_{x\to-\infty}3x^{-7}+x^2=\infty\]

Answer 12

right polynomials don't negative powers. so what would you do in this case

Answer 13

\[\lim_{x\to-\infty}3x^{-7}=0\]

\[\lim_{x\to-\infty}x^2=\infty\]

Answer 14

\[\lim_{x \rightarrow \infty}(\frac{3}{x^7}+x^2)=\lim_{x \rightarrow \infty}\frac{3+x^9}{x^7}=\lim_{x \rightarrow \infty}\frac{\frac{1}{x^7}}{\frac{1}{x^7}} \frac{3+x^9}{x^7}\]
\[=\lim_{x \rightarrow \infty}\frac{\frac{3}{x^7}+\frac{x^9}{x^7}}{1}=\lim_{x \rightarrow \infty} \frac{0+\infty}{1}=\infty\]

Answer 15

and of course i didn't mean to leave that lim x->infty there after i took the limit of each part

Answer 16

I am sorry I am not getting it, thank you for your help