1. Sepecial limit

Calculate :

2. Originally Posted by dhiab
Calculate :

$\sqrt[4]{x^4+1}=\sqrt[4]{x^4(1+\frac{1}{x^4})}=x\sqrt[4]{1+\frac{1}{x^4}}$

since $\sqrt[4]{1+\frac{1}{x^4}}\to{1}$ as $x\to{+\infty}$, by substitution we have $+\infty(1)-(+\infty)=huh?$

3. if you're trying to say that $\infty-\infty=0,$ your solution is wrong.

4. Originally Posted by Krizalid
if you're trying to say that $\infty-\infty=0,$ your solution is wrong.

Well, I admit that I don't have a good grasp on limits of this nature, but I do know that

$\lim_{x\to{c}}f(x)*g(x)=[\lim_{x\to{c}}f(x)][\lim_{x\to{c}}g(x)]$

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?

5. Multlpyling and dividing by the conjugate doesn't seem to help.

$\lim_{x \to \infty} (\sqrt[4]{x^{4}+1} - x) \frac {\sqrt[4]{x^{4}+1} + x}{\sqrt[4]{x^{4}+1} + x}$

$= \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}$

but then the limit is indeterminate $\big(\frac {0}{0}\big)$, which can been seen if you divide the top and bottom by $x^{2}$

I guess you could then use L'Hospital's rule, but it gets messy.

6. Originally Posted by Random Variable
Multlpyling and dividing by the conjugate doesn't seem to help.

$\lim_{x \to \infty} (\sqrt[4]{x^{4}+1} - x) \frac {\sqrt[4]{x^{4}+1} + x}{\sqrt[4]{x^{4}+1} + x}$

$= \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}$

but then the limit is indeterminate $\big(\frac {0}{0}\big)$, which can been seen if you divide the top and bottom by $x^{2}$

I guess you could then use L'Hospital's rule, but it gets messy.
Applying L'Hopital

$= \lim_{x \to \infty} \frac {\sqrt{x^{4}+1} - x^{2}}{\sqrt[4]{x^{4}+1} + x}$

$=\frac{x^3(x^4+1)^{-3/4}-2x}{x^3(x^4+1)^{-3/4}+1}$

$=\frac{(x^4+1)^{-3/4}-2x}{(x^4+1)^{-3/4}+1}$

$=\frac{(x^4+1)^{-3/4}}{(x^4+1)^{-3/4}+1}\to{1}$

and

let $u=(x^4+1)$

$=\frac{(u)^{-3/4}}{(u)^{-3/4}+1}\to{1}$

someone that knows what they're doing should finish this. I suck

7. I feel like a total idiot for not seeing an easy way to do this. The limit is obviously 0, though.

8. here's my effort
$\lim_{x\to\infty}{\sqrt[4]{x^4+1}-x}$
= $\lim_{x\to\infty}{\sqrt[4]{x^4+1}-x\left(\frac{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}{\sqrt[4]{(x^4+1)^3}+\sqrt[4]{x^4+1}*x^2+\sqrt{x^4+1}*x+x^3}\right)}$
$=\lim_{x\to\infty}\frac{1}{x^3\sqrt[4]{(1+\frac{1}{x^4})^3}+x^3\sqrt[4]{1+\frac{1}{x^4}}+x^4\sqrt[4]{1+\frac{1}{x^2}}+x^3}$

now divide by x^4 and puting the limit it will be

$\frac{0}{0+0+1+0} = 0$

correct me if i made a mistake.

9. I'm just gonna stay on my level for now. i thank you all.

10. Originally Posted by dhiab
Calculate :
Substitute $x = \frac{1}{t}$ and consider the limit $t \rightarrow 0$.

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.

11. Originally Posted by dhiab
Calculate :
$(x^4+1)^{1/4}-x=x(1+1/x^4)^{1/4}-x=x\left(1+\frac{1}{4x^4}+O(x^{-8})\right)-x=\frac{1}{4x^3}+O(x^{-7})$

Hence:

$
\lim_{x\to \infty}\left((x^4+1)^{1/4}-x\right)=0
$

Supporting evidence for this can be found by brute computation.

CB

12. I would just like to say that I appreciate how helpful Captain Black and Mr.fantastic have been. I was racking my brain for hours over this one. I knew that there were easier ways.

13. Originally Posted by CaptainBlack
$(x^4+1)^{1/4}-x=x(1+1/x^4)^{1/4}-x=x\left(1+\frac{1}{4x^4}+O(x^{-8})\right)-x=\frac{1}{4x^3}+O(x^{-7})$

Hence:

$
\lim_{x\to \infty}\left((x^4+1)^{1/4}-x\right)=0
$

Supporting evidence for this can be found by brute computation.

CB
If you are not happy with big-O notation you can use the remainder for the Taylor expansion instead.

CB

14. Another resolution

Hello Here is another resolution :
On using this identity :
and put :

they acquire :
they will have as zero