Question

HELLP

Answer 1

\[\lim(3e^xcos x)\]
\[x \rightarrow \pi/2\]

Answer 2

that x -> pi/2 is under the lim

Answer 3

@zepdrix by any chance are you familiar with this kinda question too :(

Answer 4

it says find the limit by using direct substitution, but I have no idea how to do that with this type.

Answer 5

Hmm we're not approaching any trouble spots, so we can plug our value directly in.\[\large 3e^{\pi/2}\cos(\pi/2)\]

Answer 6

Limits can be a little weird.
If we're not getting an indeterminate form, we're allowed to plug the value directly in as we did here.

Answer 7

what is e??? LOL

Answer 8

You silly girl +_+

\(\large e \approx 2.71828\)
It's an irrational number like \(\large \pi\).

Answer 9

Don't worry about the exponential so much though.
First simplify the trig function.

\(\large \cos(\pi/2)=?\)

Answer 10

I know it looks a little ugly, but believe it or not, this part is already simplified.
\[\large e^{\pi/2}\]

Answer 11

ahh!! der da der... LOL sorry! okay it's 0 right?

Answer 12

Yesss, good good.

Answer 13

So that makes the entire thing zero since we're multiplying, right?

Answer 14

Yep! :) wow that was actually easy, it looked treacherous before though.. xD

Answer 15

heh

Answer 16

thanks :)

Answer 17

wait so I have a question about this, if the question said solve it algebraically would I directly plug in what x was under the limit?

Answer 18

@zepdrix

Answer 19

If it asks you to solve it algebraically, then it means that you are unable to directly plug in the limit value because of some issue.

Example, if the problem had instead been:\[\Large \lim_{x\to \pi/2}\frac{3e^x}{\cos x}\]

We would see that by direct substitution, we end up dividing by 0.
It would give us an indeterminate form, so we can't directly substitute.
We would have to use some algebraic methods to change this thing around, simplify it in some way, BEFORE we can substitute in the limit value.

Answer 20

wait okay so for ex I have a question that say to find the limit algebraically and it's lim x^2-81/x-9 does it want me to break it down to (x+9)(x-9)/x-9 ,
x->9 cancel out the x-9's and then plug the 9 into (x+9) to
get 18?

Answer 21

woah the question really screwed up..

Answer 22

Yes very good! :)
You saw that by direct substitution we had a problem, it was giving us 0/0.

So you jimmy'd with it a bit doing some algebra, and got it to a point where plugging in x=9 no longer gave you a problem.

Answer 23

\[\lim_{x \rightarrow 9} x^2-81/x-9\]

Answer 24

AWESOME!! okay cool I'm finally starting to understand this xD thank you!!!

Answer 25

np c: