limit

[quasar987]

I Have No Clue How To Start This One. I Tried Applying The Same Kind Of Strategy As In http://www.physicsforums.com/showthread.php?t=51562 But No Luck. Please Give Me A Hint.

\lim_{x\rightarrow 0}\frac{\sqrt[4]{1+x^2}-1}{x}

Factoring X Out Gives A \infty - \infty Undeterminate Form. The answer is 0.

[arildno]

Set:
a=\sqrt[4]{1+x^{2}}, b=1^{\frac{1}{4}}=1
Find the polynomial in a, b P(a,b) which satisfies:
(a-b)P(a,b)=a^{4}-b^{4}
In order to find P(a,b), use polynomial division on:
(a^{4}-b^{4}):(a-b)

In order then to evaluate the limit, multiply your fraction with:
1=\frac{P(a,b)}{P(a,b)}

[quasar987]

Simply amazing!

And I realize this method is the same as the one which have been advised to me for the other limit problem, but generalized. Thanks arildno !