help me limit question

Question

anonymous · Answer

$$\lim_{x \rightarrow a}(g(x))^1$$

anonymous · Answer

i mean ^-1 in the end

anonymous · Answer

$$\lim_{x \rightarrow a}g(x)=6$$

jim_thompson5910 · Answer

$$\Large \lim_{x \to a}(g(x))^{-1} = (\lim_{x \rightarrow a}g(x))^{-1}$$

jim_thompson5910 · Answer

does that help?

anonymous · Answer

so  the answer is 6?

jim_thompson5910 · Answer

no, not quite

anonymous · Answer

mmm -6?

jim_thompson5910 · Answer

$$\Large \lim_{x \to a}(g(x))^{-1} = (\lim_{x \rightarrow a}g(x))^{-1}$$

$$\Large \lim_{x \to a}(g(x))^{-1} = (6)^{-1}$$

$$\Large \lim_{x \to a}(g(x))^{-1} = ??$$

anonymous · Answer

oH!!! 1/6???

jim_thompson5910 · Answer

yep

anonymous · Answer

OHHH coold! thank yoou!! are you also good at limit graph?

jim_thompson5910 · Answer

sure, what's the question

anonymous · Answer

ok, i will type it up. its quit long

anonymous · Answer

c) Graph the function to see if it is consistent with your answers to parts (a) and (b). By graphing, find an interval for x near zero such that the difference between your conjectured limit and the value of the function is less than 0.01. In other words, find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom. What is the window?  

x=

2.9

2.99

2.999

2.9999

3.0001

3.001

3.01

3.1

f(x)
5.9 5.99 5.999 5.9999 5.99999 5.999999 6.01 6.1

$$ \lim_{x \rightarrow 3} \frac{ x^2-9 }{ x-3 }$$

jim_thompson5910 · Answer

where are parts a) and b) ?

anonymous · Answer

a and b is below: ( lim equation and decimal numbers!)

jim_thompson5910 · Answer

f(x) 5.9 5.99 5.999 5.9999 5.99999 5.999999 6.01 6.1 

is part a) right?

anonymous · Answer

yes also X=2.9 2.99 etc
I have to answer the question like this : 
<x<
<y<

jim_thompson5910 · Answer

ok what did you get for part a

anonymous · Answer

part a) i got f(x)= 5.9 5.99 5.999 etc

anonymous · Answer

a) 5.9 5.99 5.999 5.9999 5.99999 5.99999 5.999999 6.01 6.1

jim_thompson5910 · Answer

$$ f(x) = \frac{ x^2-9 }{ x-3 }$$


$$ f(2.9) = \frac{ (2.9)^2-9 }{ 2.9-3 }$$


$$ f(2.9) = \frac{ 8.41-9 }{ 2.9-3 }$$


$$ f(2.9) = \frac{ -0.59 }{ -0.1 }$$


$$ f(2.9) = 5.9$$

so that's correct

anonymous · Answer

b) $$\lim_{x \rightarrow 3} \frac{ x-^2-9 }{ x-3} = 6$$

jim_thompson5910 · Answer

$$ f(x) = \frac{ x^2-9 }{ x-3 }$$


$$ f(2.99) = \frac{ (2.99)^2-9 }{ 2.99-3 }$$


$$ f(2.99) = \frac{ 8.9401-9 }{ 2.99-3 }$$


$$ f(2.99) = \frac{ -0.0599 }{ -0.001 }$$


$$ f(2.99) = 5.99$$

that's correct too

so it looks like you know what you're doing for part a

anonymous · Answer

thank you!

jim_thompson5910 · Answer

and part b is correct because x^2 - 9 = (x-3)(x+3)

the x-3 terms will cancel and you jut plug in x = 3 to get x+3 = 3+3 = 6

anonymous · Answer

but i have no idea about graphig

jim_thompson5910 · Answer

so part a) is slowly getting close to 6 while part b shows you that the limit is exactly 6

jim_thompson5910 · Answer

they both tell the same story

jim_thompson5910 · Answer

if you were to graph $$ f(x) = \frac{ x^2-9 }{ x-3 }$$ and locate the point at x = 3, you should see that the y value is 6 and that each neighboring point is close to 6 as well

anonymous · Answer

so would it be 6<x6?

anonymous · Answer

oh no 3<x<3?

jim_thompson5910 · Answer

what did you get for your conjectured limit

anonymous · Answer

conjectured limit?

jim_thompson5910 · Answer

what limit did you get in part a

jim_thompson5910 · Answer

that's your conjectured limit

anonymous · Answer

oh i see. 6?

jim_thompson5910 · Answer

thats the actual limit

anonymous · Answer

ok hold on im taking notes!

anonymous · Answer

so 5.9 5.99 etc are called conjectured limit

jim_thompson5910 · Answer

yes pick the closest one

anonymous · Answer

6.01?

jim_thompson5910 · Answer

You can go closer

anonymous · Answer

5.999999!!

jim_thompson5910 · Answer

better

anonymous · Answer

so whats next step to graph?

jim_thompson5910 · Answer

yes, but we have to find the window first, one sec

jim_thompson5910 · Answer

hmm these instructions are strange

anonymous · Answer

i think they asking doman and range or something?

jim_thompson5910 · Answer

clearly 5.9999 is within 0.01 of the limit 6

jim_thompson5910 · Answer

so why is that a requirement

jim_thompson5910 · Answer

i guess we could try this

find a window of height 0.02 such that the graph exits the sides of the window and not the top or bottom

anonymous · Answer

mmmm im stuck!

jim_thompson5910 · Answer

well for one thing, I know that the y part of the window is 

5.99 < y < 6.01

since the height of the window must be 0.02

jim_thompson5910 · Answer

the key is to find the x part

anonymous · Answer

oh i see why x is not the same?

anonymous · Answer

dx=

2.9

2.99

2.999

2.9999

3.0001

3.001

3.01

3.1

jim_thompson5910 · Answer

well x would be near 3, not 6

so that's one reason why it's not the same

jim_thompson5910 · Answer

maybe if you did 2.999 < x < 3.001

that might work

jim_thompson5910 · Answer

I just used winplot and it seems to work out and look good

anonymous · Answer

I see. !!

anonymous · Answer

Thank you soo much for your help!!

jim_thompson5910 · Answer

you're welcome

anonymous · Answer

I was able to understand