Calculate :
Well, I admit that I don't have a good grasp on limits of this nature, but I do know that
$\displaystyle \lim_{x\to{c}}f(x)*g(x)=[\lim_{x\to{c}}f(x)][\lim_{x\to{c}}g(x)]$
and that the limit of a sum is the sum of the limits.
I know I used good math here. The only part that I was unsure of was subtraccting infinity from infinity.
Could you please enlighten me with an explaination?
Multlpyling and dividing by the conjugate doesn't seem to help.
$\displaystyle \lim_{x \to \infty} (\sqrt[4]{x^{4}+1} - x) \frac {\sqrt[4]{x^{4}+1} + x}{\sqrt[4]{x^{4}+1} + x} $
$\displaystyle = \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x} $
but then the limit is indeterminate $\displaystyle \big(\frac {0}{0}\big) $, which can been seen if you divide the top and bottom by $\displaystyle x^{2} $
I guess you could then use L'Hospital's rule, but it gets messy.
Applying L'Hopital
$\displaystyle = \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x} $
$\displaystyle =\frac{x^3(x^4+1)^{-3/4}-2x}{x^3(x^4+1)^{-3/4}+1}$
$\displaystyle =\frac{(x^4+1)^{-3/4}-2x}{(x^4+1)^{-3/4}+1}$
$\displaystyle =\frac{(x^4+1)^{-3/4}}{(x^4+1)^{-3/4}+1}\to{1}$
and
let $\displaystyle u=(x^4+1)$
$\displaystyle =\frac{(u)^{-3/4}}{(u)^{-3/4}+1}\to{1}$
someone that knows what they're doing should finish this. I suck
here's my effort
$\displaystyle \lim_{x\to\infty}{\sqrt[4]{x^4+1}-x}$
= $\displaystyle \lim_{x\to\infty}{\sqrt[4]{x^4+1}-x\left(\frac{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}\right)}$
$\displaystyle =\lim_{x\to\infty}\frac{1}{x^3\sqrt[4]{(1+\frac{1}{x^4})^3}+x^3\sqrt[4]{1+\frac{1}{x^4}}+x^4\sqrt[4]{1+\frac{1}{x^2}}+x^3}$
now divide by x^4 and puting the limit it will be
$\displaystyle \frac{0}{0+0+1+0} = 0$
correct me if i made a mistake.
Substitute $\displaystyle x = \frac{1}{t}$ and consider the limit $\displaystyle t \rightarrow 0$.
If this had been posted in the Calculus subforum you could now easily apply l'Hopital's rule.
However, without using l'Hopital's Rule, there's slightly (but not a lot) more work ..... multiply numerator and denominator by the conjugate surd (twice) and then cancel a common factor of t.
Either way, you get 0 as the answer.