Just because his class does use l'Hopital's Rule doesn't mean he shouldn't use it

SAQLAIN AHMAD

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When we are trying to solve limits involving different powers on numerator and denominator, we will just simplily divide the whole function with the highest power. So I tried to apply this method in a limit as the following
lim x-∞ (1)/(√(x^2-x)-x), diving the both the denominator and numerator with √(x^2). However the result is wrong. I know that this limit has to be solved by multilying something like (a-b)(a+b). But why can't we divide a variable? And under what situation can we divide a variable in a infinity limit?
Many thanks and happy new year!

dividing numerator and denominator by x,we get Limit=Lim x-∞ (1/x)/(√(1-(1/x))-1) Let 1/x=y. Then y-0 So L=Lim y-0 y/(√(1-y)-1).Apply L-Hospital rule,we get L=Lim y-0 1/(-1/(2*√(1-y))).Tend y to 0.So L=1/(-1/2)=-2.So ans is -2.So dividing by highest power works when numerator and denominator both tends to infinity as x-infinity
 
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dividing numerator and denominator by x,we get Limit=Lim x-∞ (1/x)/(√(1-(1/x))-1) Let 1/x=y. Then y-0 So L=Lim y-0 y/(√(1-y)-1).Apply L-Hospital rule,we get L=Lim y-0 1/(-1/(2*√(1-y))).Tend y to 0.So L=1/(-1/2)=-2.So ans is -2.So dividing by highest power works when numerator and denominator both tends to infinity as x-infinity

Thanks. What a shame that my school doesn't include L'Hospital rule in limit curriculum.
 
Thanks. What a shame that my school doesn't include L'Hospital rule in limit curriculum.
There is absolutely no reason whatsoever to use L'Hospital's rule. Why was it even mentioned? Here is you original question:
\(\displaystyle \displaystyle{{\lim _{x \to \infty }}\frac{1}{{\sqrt {{x^2} - x} - x}} = {\lim _{x \to \infty }}\frac{{\sqrt {{x^2} - x} + x}}{{ - x}} = - 2}\) SEE HERE

Both the original post and the reply are hopelessly obscure notation wise. Such messes are almost impossible to read. For example, lim(x-infinity)?? How does that mean x approaches infinity?
Look at my reply. Using simple LaTeX coding it is completely readable.

As for the other reply, not only is it hard to read it is also miss-leading. Can you see how a simple algebra problem is turned it a complicated mess?
It is simple algebra because multiplying by \(\displaystyle \frac{{\sqrt {{x^2} - x} + x}}{{\sqrt {{x^2} - x} + x}}\) is all it takes.
 
There is absolutely no reason whatsoever to use L'Hospital's rule. Why was it even mentioned? Here is you original question:
\(\displaystyle \displaystyle{{\lim _{x \to \infty }}\frac{1}{{\sqrt {{x^2} - x} - x}} = {\lim _{x \to \infty }}\frac{{\sqrt {{x^2} - x} + x}}{{ - x}} = - 2}\) SEE HERE

Both the original post and the reply are hopelessly obscure notation wise. Such messes are almost impossible to read. For example, lim(x-infinity)?? How does that mean x approaches infinity?
Look at my reply. Using simple LaTeX coding it is completely readable.

As for the other reply, not only is it hard to read it is also miss-leading. Can you see how a simple algebra problem is turned it a complicated mess?
It is simple algebra because multiplying by \(\displaystyle \frac{{\sqrt {{x^2} - x} + x}}{{\sqrt {{x^2} - x} + x}}\) is all it takes.

You just didn't understand the question.The student knows the approach which you have given.He asked when to divide by highest power.So I divided by the highest power and applied L-Hospital instead of the mthod you gave
 
You just didn't understand the question.The student knows the approach which you have given.He asked when to divide by highest power.So I divided by the highest power and applied L-Hospital instead of the mthod you gave
Why do that? He clearly says that his school does not use the L'Hospital's rule. So why confused him?
Why not show how to do it the correct way according to standard methods.
Like this: if \(\displaystyle x\to\infty\) then it is true that \(\displaystyle x>0\) therefore, \(\displaystyle \dfrac{\sqrt{x^2-x}+x}{-x}=-\left(\sqrt{1-\frac{1}{x}}+1\right)\to-2\)

Perhaps it you who lacks understanding as to what this is about?
Once again please learn LaTeX coding, it really helps in explaining problems.
 
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