Help on finding the limit

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Ryousekki

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I have this equation i need to find the limit
(√(x+3))-2/(x-1)

What is the limit if x approaches 1

The prof said the limit cant be 0 or undefined
But all the answers i come up always lead to 0 please help im stuck
 

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Ryousekki said:
I have this equation i need to find the limit
(√(x+3))-2/(x-1)
What is the limit if x approaches 1
Click to expand...
Please confirm that the question is: \(\displaystyle \mathop {\lim }\limits_{x \to 1} \left( {\sqrt {x + 3} - \frac{2}{{x - 1}}} \right)\)
 
Ryousekki said:
I have this equation i need to find the limit
(√(x+3))-2/(x-1)

What is the limit if x approaches 1

The prof said the limit cant be 0 or undefined
But all the answers i come up always lead to 0 please help im stuck
Click to expand...

Ryousekki said:
The question is
(√x+3)-2 / x-1
Click to expand...
In the future, please be sure to write expressions so that what they mean what you intend.

What you wrote first, (√(x+3))-2/(x-1), means [MATH]\sqrt{x+3} - \frac{2}{x-1}[/MATH], and has limit [MATH]-\infty[/MATH].

What you wrote this time, (√x+3)-2 / x-1, means [MATH](\sqrt{x}+3) - \frac{2}{x} - 1[/MATH], and has limit 1.

What I suspect you meant (since it has a more interesting limit) is (√(x+3)-2)/(x-1), which means [MATH]\frac{\sqrt{x+3} - 2}{x-1}[/MATH], and has limit [MATH]\frac{1}{4}[/MATH].

Communicating clearly is very important when you ask for help!
 
I admit I'm stumped on how to do [math]\lim_{x \to 1} \dfrac{\sqrt{x + 3} - 2}{x - 1}[/math] without l'Hopital's rule. Can someone start me off?

-Dan
 
This is where Jomo cracks a joke about physicists.
No offense. :)
Since x approaches one by a square root in the numerator, you can graph it, but it's 1/4, right? I am confused.
 
Last edited:
I may work on dan's problem later, but it is not obvious to me, and I have work to do right now. I think it may have to do with why L'Hospital's Rule works in the first place, but I am not at all sure.
 
topsquark said:
I admit I'm stumped on how to do [math]\lim_{x \to 1} \dfrac{\sqrt{x + 3} - 2}{x - 1}[/math] without l'Hopital's rule. Can someone start me off?

-Dan
Click to expand...
How about:

\(\displaystyle \sqrt{x+3}\) = u

x - 1 = u2 - 4

[math]\lim_{x \to 1} \dfrac{\sqrt{x + 3} - 2}{x - 1}[/math]
= [math]\lim_{u \to |2|} \dfrac{u-2}{u^2 - 4}[/math]
= [math]\lim_{u \to |2|} \dfrac{1}{u+2}[/math]
=\(\displaystyle \frac{1}{4}\)
 
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