Question

Evaluate the limit as x approaches 1 of the quotient of the square root of the quantity x squared plus 8 minus 3 and the quantity x minus 1. You must show your work to receive credit.

Answer 1

The two-sided limit does not exist.

limit as x approaches 1 of [(square root x^2+8) -3]/[x-1]
limit as x approaches 1 of [(square root 1^2+8)-3]/[1-1]
limit as x approaches 1 of [(square root 9)-3]/[0]
limit as x approaches 1 of [(3)-(3)]/[0]
limit as x approaches 1 of [0]/[0]
DNE, undefined.

Answer 2

@Luigi0210 please help!! :(

Answer 3

@Luigi0210 is there another way to evaluate this limit or is this the only way? could we find a way to not make the bottom 0?

Answer 4

Do you know L'hopital's ?

Answer 5

@ikram002p @Luigi0210 no can u please write out the steps so i can follow them :) this is an essay question so i neeed to understand every step of the problem.

Answer 6

Calc I right?

Answer 7

yes :)

Answer 8

are you there ? :)

Answer 9

@Luigi0210

Answer 10

If you don't know it then I assume they don't want you to know about it yet. So yea, when it approaches 1, it is DNE

Answer 11

no i think im supposed 2 know it

Answer 12

i just need 2 brush up on it :)

Answer 13

i know were gonna have 2 factor? i think

Answer 14

Pretty much this: http://prntscr.com/65l363

If you get 0/0, or inf/inf, then you are allowed to take the derivatives of both and try the limit again

Answer 15

ok thanks can u show the steps on here plz? :)

Answer 16

Okay so we have \(\Large \lim_{x \rightarrow 1} ~\frac{\sqrt{x^2+8}}{x-1} \) which turned out to be \(\frac{0}{0} \) . Now, according to L'hopital's, we can do this:

\(\Large \lim_{x \rightarrow 1} ~\frac{dy/dx (\sqrt{x^2+8})}{dy/dx (x-1)} \)

Answer 17

Do you know how to take the derivatives?

Answer 18

Or do you need help on that?

Answer 19

yes i need help on it :) btw thanks for the 1st step :)

Answer 20

So for \(\large \sqrt{x^2+8} \) can be rewritten as \(\large (x^2+8)^{1/2} \) .

For this one we have to use chain rule: First take the derivative of the outside, which is bringing down the 1/2 and subtracting 1.

\(\Large \color{red}{\frac{1}{2}} * (x^2+8)^{\color{red}{\frac{1}{2}-1}} \)

We get \(\Large \frac{1}{2}(x^2+8)^{\color{green}{-1/2}} \)

Now for the chain rule, take the derivative of the inside:

\(\Large \frac{1}{2} (\color{blue}{x^2+8})^{-1/2} *\color{blue}{2x+0} \)

Since dy/dx of x^2 is 2x, because we bring down the 2 and subtract 1. And the dy/dx of any constant is 0.

Make sense so far?

Answer 21

yes :)!!! perfect explanation

Answer 22

Okay good. Now, if you remember the rules of exponents, whenever it is negative, we are allowed to flip it to make it positive, so it becomes:

\(\Large \frac{2x}{2\sqrt{x^2+8}} \)

The 2's can cancel:

\(\Large \frac{x}{\sqrt{x^2+8}} \)
And we have that^

Answer 23

Now, think you can take the dy/dx of x-1?

Answer 24

wouldn't it just be 0?

Answer 25

actually the answer is 1 because of the sum rule...right?

Answer 26

Nopee, remember the x still have a power of 1.


So it would actually be: \(\large \color{red}{1} x^{\color{red}{1-1}} =1*(x^0)\)

and x^0 is 1. so: \(\Large 1*1=1 \)

The -1 however is a constant, so that is 0

Answer 27

Without LH:
\[\lim_{x \rightarrow 1}\frac{\sqrt{x^2+8}-3}{x-1} \\ \text{ you can also choose to do this without derivatives } \\ \text{ you can decide to rationalize the numerator by } \\ \text{ multiplying top and bottom by top's conjugate } \\ \lim_{x \rightarrow 1}\frac{\sqrt{x^2+8}-3}{x-1} \cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3} \\ \lim_{x \rightarrow 1}\frac{(x^2+8)-9}{(x-1) (\sqrt{x^2+8}+3)} \]

\[\text{ You will need to know how to factor a difference of squares } \\ x^2-a^2=(x-a)(x+a)\]

Answer 28

Yea, the conjugate works too.. depends which way you wanna go ^

Answer 29

ohh freckles i like that idea!!! it seems easier, luigi do we have a long way 2 go?

Answer 30

Not really, now all we have is:

\(\Large \lim_{x \rightarrow 1} \frac{x}{\sqrt {x^2+8}} -3 \) :P

Answer 31

hey what's that -3 there for?

Answer 32

Oh, whoops, yea, sorry about that xD

Answer 33

Ignore that -3. it turns to 0

Answer 34

ok freckles can u please show the next step after limit as x approaches 1 of (x^2+8)-9/(x-1)[(square root x^2+8) +3]

Answer 35

okay then luigi do i just plug in the 1 now?

Answer 36

Yup, it should give you an answer.. if not, well then L'Hop again if it's 0/0 xD

Answer 37

i think it gives: 1/3

Answer 38

is it correct?:)

Answer 39

It most definitely is :P

Answer 40

THANKU!!!!!!!!!!!!!!!!!!!!! LIFE SAVERRRRRRRR

Answer 41

i have 6 more questions, are you available to help? :)

Answer 42

You're welcome xD

And sure, I can try :P

Answer 43

I know you got the answer...But you asked what happens after that
Well the hint was to fact the difference of squares.
The difference of squares was on top if you simplified the top.
That is the top should be x^2-1
which equals (x-1)(x+1)
Now you will be able to cancel a common factor from top and bottom
That is you should have:
\[
\lim_{x \rightarrow 1}\frac{\sqrt{x^2+8}-3}{x-1} \\ \\ \text{ you can decide to rationalize the numerator by } \\ \text{ multiplying top and bottom by top's conjugate } \\ \lim_{x \rightarrow 1}\frac{\sqrt{x^2+8}-3}{x-1} \cdot \frac{\sqrt{x^2+8}+3}{\sqrt{x^2+8}+3} \\ \lim_{x \rightarrow 1}\frac{(x^2+8)-9}{(x-1) (\sqrt{x^2+8}+3)}
\\ \lim_{x \rightarrow 1} \frac{x^2-1}{(x-1)(\sqrt{x^2+8}+3)} \\ \lim_{ x \rightarrow 1}\frac{(x-1)(x+1)}{(x-1)(\sqrt{x^2+8}+3)} \\ \lim_{x \rightarrow 1} \frac{x+1}{\sqrt{x^2+8}+3} \]
Now you can replace x with 1.