limit question. No L'Hospital rule..

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limit question. No L'Hospital rule..

Mathematics
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\[\LARGE \lim_{x \rightarrow a}\frac{x^\sqrt{2}-a^\sqrt{2}}{x-a}\]
I would consider rationalizing the numerator AND denominator first.
sqrt2 power of x and a

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Other answers:

Never mind. Factor the numerator.
you there?
@cinar: If that was your answer you are incorrect. I should mention.
|dw:1355956723907:dw|
the answer is \[\sqrt{2}a^{\sqrt{2}-1}\]if L'Hospital rule is allowed, but without using it, I have no idea..
This is the (less often seen) version of the Limit Definition of a Derivative,\[\large f'(x)=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]So identity f(x), then just take it's derivative.
I am not looking for derivative of it, it is limit question..
This limit represents the LIMIT DEFINITION OF A DERIVATIVE. You are not actually expected to evaluate the limit. You're suppose to recognize that it represents a derivative.
Example: Evaluate the limit:\[\large \lim_{h \rightarrow 0}\frac{(x+h)^2-x^2}{h}\] You could do the work simplifying it down, Or you can recognize that this is the DERIVATIVE of\[\large f(x)=x^2\]
\[ f'(x)=\frac{a^{\sqrt2}+\sqrt2 x^{\sqrt2-1} (x-a)-x^{\sqrt2}}{(a-x)^2}\]
...?
\[f(x)=x^{\sqrt2} ?\]
Yes, good.
I see now..
thanks..
\[\large f'(\color{red}{a})=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}\]Ah sorry I made a small typo,
but we cannot simplify it can we?
So that's where the a in coming from in the answer i guess :D
You just need to recognize that,\[\large f(x)=x^{\sqrt2}\]And that this limit represents,\[\large f'(a)\] So simply take the derivative of f(x), then plug in a! :)
cool..
\[\large \lim_{h \rightarrow 0}\frac{(x+h)^2-x^2}{h}=2x ?\]

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