Question

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

Answer 1

Post them! :)

Answer 2

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

Answer 3

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

Answer 4

oh you did the first one ok

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

and where a=9

Answer 12

no I have not learned that yet.

Answer 13

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

Answer 14

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

Answer 15

ln(9) is close to 2

Answer 16

that is pretty close

Answer 17

2.2 is a better number

Answer 18

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

Answer 19

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

Answer 20

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.

Answer 21

I would make a piecewise function here

Answer 22

well you know what lets not do that

Answer 23

I don't know what a piecewise function is.

Answer 24

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

Answer 25

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

Answer 26

25

Answer 27

that's right!

Answer 28

you know how i came up with my function?

Answer 29

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.

Answer 30

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

Answer 31

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

Answer 32

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

Answer 33

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

Answer 34

so I knew 8+17=25

Answer 35

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

Answer 36

since 8+17 is 25

Answer 37

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

Answer 38

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.

Answer 39

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

Answer 40

people are addicted to this site

Answer 41

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

Answer 42

have a good day

Answer 43

and thanks and welcome lol