Question

How would I evaluate the following Limit?

Answer 1

\[\Huge Evaluate \lim_{t \rightarrow -2}(t+1)^9(t^2-1)?\]

Answer 2

That \(\ \Huge ^9 \) confuses me

Answer 3

Trivially, I think. One moment.

Answer 4

\(f(t)=(t+1)^9\) and \(g(t)=(t^2-1)\) are both continuous functions in \(\mathbb{R}\), so we have:

\[\lim_{t \to -2}f(t)g(t)=f(-2)g(-2)\]

Answer 5

just plug in -2 for t like any normal equation

polynomials have domains of all reals, meaning any number can be put in.

-2 is no exception. just plug it in, and solve

Answer 6

What about that 9?

Answer 7

^i just said that, Funinabox!

Answer 8

There's no g(x)...

Answer 9

Yes, there is.. I defined them in the first sentence.

Answer 10

This works because \[\lim_{x \to a}f(x)=f(a)\] for a continuous function \(f\).

Answer 11

So how would I figure out \(\ \Huge (-1)^9? \) I always get that confused...

Answer 12

And (-2)^2 is 4 right?

Answer 13

Silly Questions, I know.

Answer 14

Just punch it in your calculator.

Answer 15

\((-1)^n\) is either \(1\) or \(-1\). If \(n\) is even, \((-1)^n=1\). If \(n\) is odd, \((-1)^n=-1\).

Answer 16

Yes, \((-2)^4=(-2)(-2)=4\).

Answer 17

So, \(\ \Huge (-1)^9=-1 ?\)

Answer 18

Yup, because \(9\) is odd.

Answer 19

Thanks! This makes so much more sense now!

Answer 20

Here's essentially "what's happening":
\[(-1)^9=(-1)^{8+1}=(-1)^{8}(-1)^{1}=(-1)^{2\cdot 4}(-1)^{1}=((-1)^{2})^{4}(-1)^{1}=(1)^{4}(-1)^{1}=-1\]

Answer 21

And you're welcome! :D

Answer 22

Just plug in your -2 into t and solve it.
Whenever you have a limit, ALWAYS try plugging in the value first. Always. If you don't have a zero at the bottom then you are GOOD:)