what is this limit
\[\lim_{x \rightarrow \pi/2}(xcosx)\]

- anonymous

what is this limit
\[\lim_{x \rightarrow \pi/2}(xcosx)\]

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- anonymous

\[\lim_{x \rightarrow \pi/2}(xcosx)\]

- Mertsj

0

- anonymous

mertsj do you mind demonstrating that please?

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## More answers

- Mertsj

pi/2times cospi/2=pi/2times 0 = 0

- anonymous

ok thanks would you mind helping me with this one also

- anonymous

\[\lim_{x \rightarrow -7}(x^2 + 10x + 21/x + 7)\]
i've got -98/0. i dont know what to do from there

- Mertsj

You can't divide by 0 so divide numerator so another approach is needed.

- Mertsj

Factor and reduce first

- anonymous

ok i was thinking along the lines as approach -7 from left and right

- anonymous

(x+7)(x+3)/(x+7)...so the x + 7 cancels and im left with x + 3

- Mertsj

yes

- Mertsj

Now replace the x with -7

- anonymous

i see.

- anonymous

i have this one that is somewhat similar but you end up with 0/0
\[\lim_{x \rightarrow 1}(x^2 - 4x + 3/ x^2 + 2x -3)\]

- anonymous

same approach again right? factor and reduce?

- Mertsj

yes

- anonymous

(x-3)(x-1)/ (x-3)(x+1)...?

- anonymous

is that the right structure

- Mertsj

|dw:1326943979278:dw|

- anonymous

oh wow im getting this thanks for the help. would you be willing to continue working with me? i have about 8 more

- Mertsj

I am willing but I might be a little distracted as my daughter is here and we are visiting.

- anonymous

ok no problem

- anonymous

\[\lim_{x \rightarrow -10}(x^2 - 17x + 70/ x + 10)\]

- anonymous

when i attempt this...

- anonymous

i get 140/0...so i take the factor and reduce approach

- anonymous

and..

- Mertsj

Yes factor and reduce

- Mertsj

|dw:1326944396601:dw|

- anonymous

(x+10)(x+7)/ (x+10)....x+7...-10+7= -3?

- anonymous

i keep getting my signs wrong when i factor

- Mertsj

It won't reduce. Did you copy it right?

- anonymous

why couldnt it have been (x+10)(x+7)...it will also give you 70 right

- Mertsj

Middle term is negative

- anonymous

yeah thats right i remember now

- Mertsj

Did you copy it right?

- anonymous

what do you mean

- anonymous

it does not exist

- anonymous

i see that

- Mertsj

Perhaps you wrote the problem down incorrectly

- anonymous

yeah now i understand

- anonymous

i have this next one...

- Mertsj

Did you get that last one?

- anonymous

yeah

- Mertsj

What did you get?

- anonymous

\[\lim_{x \rightarrow 2}(x^2 - 4/2x^2 -5x + 2)\]

- anonymous

I got DNE (does not exist) my software accepted the answer

- anonymous

how exactly would you factor the denominator in the previous problem. it is set up strangely for me

- Mertsj

That one should factor and reduce.

- Mertsj

(2x-1)(x-2)

- anonymous

ok i was confused because i thought somehow it must add up to -5

- Mertsj

So you end up with (x+2)/(2x-1)

- anonymous

gotcha

- anonymous

next is...

- anonymous

\[\lim_{x \rightarrow 4}(\sqrt{b+0}-2/b-4)\]

- anonymous

0/0

- anonymous

do you still factor here? doesnt look like it

- Mertsj

It won't factor but why does it have sqrt(b+0) since that is just sqrt(b)?

- anonymous

its sqrt (b +0)...i would guess you do 4 +0 then squared root your answer then subtract 2

- Mertsj

But the denominator will be 0 and that is undefined.

- anonymous

yeah i got 0/0 i was thinking theres something we could do next

- Mertsj

multiply the numerator and denominator by (sqrt(b)+2)

- anonymous

if the numerator and denominator are both zeros. wouldnt mulitplying bring us back to square 1 with 0/0

- Mertsj

|dw:1326945956921:dw|

- anonymous

i see

- anonymous

|dw:1326946154087:dw|

- anonymous

the limit is as x approaches -9

- Mertsj

Multiply numerator and denominator by 9x

- Mertsj

|dw:1326946478456:dw|

- Mertsj

lim as x approaches -9 is -1/81

- anonymous

thanks im almost through

- anonymous

Let 11xâˆ’36â‰¤f(x)â‰¤x2+5xâˆ’27 . Use the Squeeze Theorem to determine limxâ†’3f(x) =

- Mertsj

ok but I still want to revisit that DNE one

- anonymous

ok when we finish we can go back

- anonymous

this one that i posted is a new concept so its unlike what we have been practicing so im clueless

- Mertsj

I see that. I have never heard of the squeeze theorem but I would assume that we should evaluate both expressions when x is 3

- anonymous

i got 33 and -3 respectively. what do i do with these two numbers to come up with a final answer

- Mertsj

You should get -3 for both of them and that is the limit. I just read it on the internet.

- Mertsj

redo the one you got the 33 for

- anonymous

yeah you're right i entered it incorrectly in my calculator

- anonymous

|dw:1326947257959:dw|
as x approaches 0

- Mertsj

I'm not sure about this one. Unless you use a reduction formula of some sort.

- anonymous

ok

- Mertsj

Have you tried typing it into wolframalpha?

- TuringTest

L'hospital does this one easily

- anonymous

|dw:1326947625929:dw|

- Mertsj

Well let's see it.

- anonymous

as x approaches 0

- Mertsj

I'll turn you over to Turing now.

- anonymous

thanks for all the help Mertsj I really appreciate it

- Mertsj

yw

- Mertsj

Ask turing about that other one also

- anonymous

ok I hope to see you around again. Enjoy

- TuringTest

@mertsj, yes, well, done
@Jinnie: do you know L'hospital's rule?
derivative of top and bottom gives \[\frac{9\cos(9x)}{1}\]

- Mertsj

Thanks

- anonymous

hey turningtest thats my first time hearing about that rule

- TuringTest

can you take the derivative of sin(9x) ?

- anonymous

no i dont know how

- TuringTest

Then you can't use L'hospital, which unfortunately means I can't help you as well as I hoped.
I can tell you that \[\lim_{x \rightarrow 0}\frac{\sin x}{x}=1\]but proof of that are bit complicated without L'hospital's rule.
All I can suggest is to try to use that fact to help you.
anyway the answer is that the limit in your question is 9

- anonymous

ok, would you be willing to do several more of these and I ask my professor about this rule next class?

- TuringTest

have you learned derivatives yet at all?

- anonymous

nah we just started school last week

- anonymous

|dw:1326948473758:dw|
lim as x approaches 0.
this is all i need help with

- TuringTest

then all I can do is give answers without explanation, because the way I solve these is with a method you don't know.
If you just want the answer I refer you to wolfram alpha as mertsj did.
Here is your question posted there:
http://www.wolframalpha.com/input/?i=limit+x+to+0+sin%289x%29%2Fx
learn to use this site, but it will not help you understand, so don't rely on it.

- TuringTest

this last one is 1/3
try typing it into wolfram to see if I'm right

- anonymous

alright thanks for the help

- anonymous

im done thanks to all

- anonymous

good night hope to see everyone again

- TuringTest

goodnight :)

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