To be computed ( source : Larson's Calculus) $\lim _{x\rightarrow\infty} \frac {x+1} {(x^2 +1)^{1/3}}$
I tried what follows, but my answer is not validated by the limit calculator I've used ( i.e. E-Math-Help, which gives : $\lim = 1$ )
$$ \lim _{x\to\infty} \frac {x+1} {(x^2 +1)^{1/3}} $$
$$ \lim _{x\to\infty} \frac {x^{3/3} ( 1+ 1/x)} {((x^2) (1+ 1/x^2))^{1/3}} $$
$$ \lim _{x\to\infty} \frac {x^{3/3} ( 1+ 1/x)} {(x^{2/3}) (1+ 1/x^2)^{1/3}} $$
$$ \lim _{x\to\infty} \frac {x^{1/3} ( 1+ 1/x)} { (1+ 1/x^2)^{1/3}} $$
Since the first factor of the denominator goes to infinity as $x$ goes to infinity ( while the second goes to $1$) and since the denominator goes to $1$ , I would like to conclude that :
$\lim _{x\rightarrow\infty} \frac {x+1} {(x^2 +1)^{1/3}}= \infty $
Where does my calculation go wrong?
Link to E-Math-Help's solution : https://www.emathhelp.net/calculators/calculus-1/limit-calculator/?f=%28+x%2B1%29%2F%28x%5E3+%2B1%29%5E%281%2F3%29&var=&val=inf&dir=
Your answer is correct. But the problem in the E-Math solution is a different one with the $x^3$ while yours is $x^2$.