Does anyone know how to do limits? I have a few homework questions I can't figure out.

- anonymous

Does anyone know how to do limits? I have a few homework questions I can't figure out.

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- jamiebookeater

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- myininaya

Post them! :)

- anonymous

I figured out the answer to the first question. I got the answer 6.

- myininaya

\[\lim_{x \rightarrow 9}\frac{x-9}{\sqrt{x}-3}\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- myininaya

oh you did the first one ok

- anonymous

The rest I am clueless on. I don't even know where to start.

- myininaya

\[\lim_{x \rightarrow 0^+}\frac{1}{\sqrt[10](x)}\]
first of all the limit is not 2

- myininaya

plug in even closer values to the right of 0 and you will see the number is getting really freaking huge
so the limit is infinity
also 1/(x^(1/10)) has a vertical asymptote at x=0
so we know it goes to infinity from the right

- anonymous

Okay great, I was on the right track because I kept getting a large number also.

- myininaya

the 2nd one is ln(9) because i remember
\[f'(9)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)\]
where
\[f(x)=\frac{a^x}{\ln(a)} \]
But i'm not sure if you know this yet

- myininaya

\[f'(x)=\frac{1}{ \ln(a)} \lim_{x \rightarrow 0}\frac{9^x-9^0}{x-0}=\ln(9)*\]

- myininaya

and where a=9

- anonymous

no I have not learned that yet.

- myininaya

ok then just plug in values really close to 0 on both sides and see if it gets closer to anything

- anonymous

I am getting close to 2. Am I way off? I am not sure if I am calculating correct.

- myininaya

ln(9) is close to 2

- myininaya

that is pretty close

- myininaya

2.2 is a better number

- anonymous

yes, I had 2.1972. We have only had whole number answers in our homework so far.

- myininaya

well limits don't have to be whole numbers
the numerical approach is okay but it only gives you approximations
so if you want to put 2 i think that is close enough

- anonymous

Thank you so much! I think I kind of knew what I was doing on these problems but didn't trust myself. Can you go over the 4th question with me.

- myininaya

I would make a piecewise function here

- myininaya

well you know what lets not do that

- anonymous

I don't know what a piecewise function is.

- myininaya

lets just say this
\[f(x)=\frac{(x-8)(x+17)}{x-8}=\frac{x^2+17x-8x- 136}{x-8}=\frac{x^2+9x-136}{x-8}\]

- myininaya

f is not defined at x=8 but
tell me what happens as x->8 ?

- anonymous

25

- myininaya

that's right!

- myininaya

you know how i came up with my function?

- anonymous

So would I write down the function exists? I don't understand how to answer what they are asking. And yes please explain how you came up with the function.

- myininaya

they want you to find a function f(x)=blah such that
f(8) does not exist and as x->8, f->25

- myininaya

so I knew I wanted this from the f(8) doesn't exist part
\[f(x)=\frac{something}{x-8}\]

- myininaya

but then i also since i wanted the limit to exist at x=8
so that means i wanted the x-8 on bottom to cancel out with one on top
\[f(x)=\frac{(x-8)(something)}{x-8}\]
but we still need a little more something on top since we don't just want 1 because we want as x->8, f->25

- myininaya

so since the (x-8)/(x-8)->1 as x->8
we only need that something to go to 25 as x->8

- myininaya

so I knew 8+17=25

- myininaya

if x->8 then x+17->25

- myininaya

since 8+17 is 25

- myininaya

\[f(x)=\frac{(x-8)(x+17)}{x-8}\]

- anonymous

I would have never figured that out. Thank you so much. And your explanation is awesome! You are one smart cookie! Too bad I couldn't have you tutor me for the res of the semester. It's very hard taking a math class online.

- myininaya

i bet it is
just keep coming back to openstudy and i or someone else will be here to help you out

- myininaya

people are addicted to this site

- anonymous

Well you are wonderful! You made my day! Thank you again!

- myininaya

have a good day

- myininaya

and thanks and welcome lol

Looking for something else?

Not the answer you are looking for? Search for more explanations.