## anonymous 3 years ago limit question. No L'Hospital rule..

1. anonymous

$\LARGE \lim_{x \rightarrow a}\frac{x^\sqrt{2}-a^\sqrt{2}}{x-a}$

2. anonymous

I would consider rationalizing the numerator AND denominator first.

3. anonymous

sqrt2 power of x and a

4. anonymous

Never mind. Factor the numerator.

5. anonymous

@cinar .

6. anonymous

you there?

7. anonymous

@cinar: If that was your answer you are incorrect. I should mention.

8. anonymous

|dw:1355956723907:dw|

9. anonymous

the answer is $\sqrt{2}a^{\sqrt{2}-1}$if L'Hospital rule is allowed, but without using it, I have no idea..

10. zepdrix

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.

11. anonymous

I am not looking for derivative of it, it is limit question..

12. zepdrix

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.

13. zepdrix

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$\large f(x)=x^2$

14. anonymous

$f'(x)=\frac{a^{\sqrt2}+\sqrt2 x^{\sqrt2-1} (x-a)-x^{\sqrt2}}{(a-x)^2}$

15. zepdrix

...?

16. anonymous

$f(x)=x^{\sqrt2} ?$

17. zepdrix

Yes, good.

18. anonymous

I see now..

19. anonymous

thanks..

20. zepdrix

$\large f'(\color{red}{a})=\lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a}$Ah sorry I made a small typo,

21. anonymous

but we cannot simplify it can we?

22. zepdrix

So that's where the a in coming from in the answer i guess :D

23. zepdrix

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! :)

24. anonymous

cool..

25. anonymous

$\large \lim_{h \rightarrow 0}\frac{(x+h)^2-x^2}{h}=2x ?$