1. ## Undifined limit problem

Hey, i have this question i can't figure out.
It's
Use the identity (a-b)(a+b)=a^2-b^2 and the arithmetic of limits to evaluate

lim root(4+5x) - 5/
x approaches 1 x-1

So plug in 1, it's undefined. I tried a few things, rationalizing the denominator, timesing by the conjugate, to try and get rid of the 0 on the bottom but didnt really get anywhere.
I'm not really sure what to do next.
Thanks in advanced for whoever helps.

2. Originally Posted by Daniiel
Hey, i have this question i can't figure out.
It's
Use the identity (a-b)(a+b)=a^2-b^2 and the arithmetic of limits to evaluate

lim root(4+5x) - 5/
x approaches 1 x-1

So plug in 1, it's undefined. I tried a few things, rationalizing the denominator, timesing by the conjugate, to try and get rid of the 0 on the bottom but didnt really get anywhere.
I'm not really sure what to do next.
Thanks in advanced for whoever helps.
1. I assume that you mean: $\lim_{x\to 1}\left(\dfrac{\sqrt{4+5x}-5}{x-1}\right)$

2. If so: $\lim_{x\to 1}\left(\dfrac{\sqrt{4+5x}-5}{x-1}\right) = \dfrac{\lim_{x\to1}(\sqrt{4+5x}-5)}{\lim_{x\to 1}(x-1)} = \dfrac{3-5}{0}$

That means the limit doesn't exist.

3. Originally Posted by earboth
1. I assume that you mean: $\lim_{x\to 1}\left(\dfrac{\sqrt{4+5x}-5}{x-1}\right)$

2. If so: $\lim_{x\to 1}\left(\dfrac{\sqrt{4+5x}-5}{x-1}\right) = \dfrac{\lim_{x\to1}(\sqrt{4+5x}-5)}{\lim_{x\to 1}(x-1)} = \dfrac{3-5}{0}$

That means the limit doesn't exist.
The only thing I would object to is writing that the limit equals that. IF the limit of those things existed, then you would say they are equal. Since the limit does NOT exist, they are not equal.

4. Thanks guys,
But it says to evaluate using that identity, and we've been taught a few ways to deal with undefined limits to see the possibilities but they haven't really worked for this one.
It's a pain in the butt, my tutor was kind of stumped on this one

5. Originally Posted by Daniiel
Hey, i have this question i can't figure out.
It's
Use the identity (a-b)(a+b)=a^2-b^2 and the arithmetic of limits to evaluate

lim root(4+5x) - 5/ <<<<<< typo?
x approaches 1 x-1

So plug in 1, it's undefined. I tried a few things, rationalizing the denominator, timesing by the conjugate, to try and get rid of the 0 on the bottom but didnt really get anywhere.
I'm not really sure what to do next.
Thanks in advanced for whoever helps.
If and only if the limit reads:

$\lim_{x\to1} \left(\dfrac{\sqrt{4+5x}-3}{x-1}\right)$ you'll get:

$\lim_{x\to1} \left(\dfrac{\sqrt{4+5x}-3}{x-1}\right) = \lim_{x\to1} \left(\dfrac{(\sqrt{4+5x}-3) \cdot (\sqrt{4+5x}+3)}{(x-1)(\sqrt{4+5x}+3)}\right)$ $=$ $\lim_{x\to1} \left(\dfrac{4+5x-9}{(x-1)(\sqrt{4+5x}+3)}\right) =$ $\lim_{x\to1} \left(\dfrac{5(x-1)}{(x-1)(\sqrt{4+5x}+3)}\right) = \left(\dfrac{5}{\lim_{x\to1}(\sqrt{4+5x}+3)}\right ) = \frac56$

6. Oh thanks alot, it was a typo
thats how it was written in my book aswell haha
how did you know it was - 3?

7. Originally Posted by Daniiel
...
x approaches 1

So plug in 1, it's undefined.
how did you know it was - 3?
I guessed the quotient should become $\frac 00$ ...