Here's the question you clicked on:
Christos
Solve the limit limit(x-->+infinity) sqrt((x-3)/x)
the square root grows slower so the limit will be zero
Divide both numerator and denominator by x, to solve it algebraically.
I am sorry the actual epression is sort(x-3/x)
\[\lim_{x\to \infty}\sqrt{\frac{x-3}{x}}\]?
then it is one since \(x-3\) and \(x\) grow at the same rate
think what it would be like if \(x\) was \(1,000\) you would have \[\sqrt{.993}\]
\[\Large \lim_{x\to \infty}\sqrt{\frac{x-3}{x}} = \sqrt {\lim_{x\to \infty}\frac{x-3} {x}}\]
bro I am trying to solve this limit actually http://screencast.com/t/Yoo5IO0bRPjD And I ended up in the expression above, but what you give me isnt the final answer, something must be wrong :(
@Christos that's a completely different expression... break it up using limit laws.
ohh let me try hold on
What do you mean by limit laws?
\[\Large = \lim_{x \rightarrow \infty} \sqrt{x^2 - 3x} -\lim_{x \rightarrow \infty} x\]
so thats infinity - infinity - infinity = +infinity?
From a graph it looks like it'll be -1.5, I'm just trying to find a way to show that algebraically.
It's ok man dont brother I found it thank you
Yeah i'm glad i didn't try to do it myself... wolfram gives a looooong solution: http://www3.wolframalpha.com/Calculate/MSP/MSP35891gdgbb0f2cc1ecie00005387i303c364idgg?MSPStoreType=image/png&s=42&w=433&h=1816