<s>[INST] Computing limit of $\sqrt{n^2+n}-\sqrt[4]{n^4+1}$ I have tried to solve this using conjugate multiplication, but I got stuck after factoring out $n^2$.
$\begin{align}
\lim_{n\rightarrow\infty}\dfrac{n^2+n-\sqrt{n^4+1}}{\sqrt{n^2+n}+\sqrt[4]{n^4+1}}
&=\lim_{n\rightarrow\infty}\dfrac{n(1+\dfrac{1}{n}-\sqrt{1+\dfrac{1}{n^4}})}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}\\&
=\lim_{n\rightarrow\infty}\dfrac{n+1-n\sqrt{1+\dfrac{1}{n^4}}}{\sqrt{1+\dfrac{1}{n}}+\sqrt[4]{1+\dfrac{1}{n^4}}}
\end{align}$
Given that $\dfrac{1}{n}$ tends to $0$ (so denominator is 2), can I reduce $n$ and $-n\sqrt{1+\dfrac{1}{n^4}}$ and say that the limit is $\dfrac{1}{2}$?
I mean $\dfrac{1}{n^4}$ tends to $0$, so $\sqrt{1+\dfrac{1}{n^4}}$ tends to $1$ and in this case $n-n\sqrt{1+\dfrac{1}{n^4}}$ can be simplified to $n-n$.
Solution given in my book uses conjugate multiplication twice to get rid of all the roots in nominator, but I am curious if my answer is correct or my teacher will tell me that simplifying the way I did it is incorrect.
 [/INST] You need to show that $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1{n^4}}\right)=0$. You cannot simply use that $\lim\limits_{n\to\infty}\sqrt{1+\frac1{n^4}}=1$ since $\lim\limits_{n\to\infty}\sqrt{1+\frac1n}=1$, yet $\lim\limits_{n\to\infty}\left(n-n\sqrt{1+\frac1n}\right)=-\frac12$.

If you are looking for an alternate approach, note that $a^4-b^4=(a-b)\left(a^3+a^2b+ab^2+b^3\right)$. Therefore,
$$
\begin{align}
\sqrt{n^2+n}-\sqrt[4]{n^4+1}
&=\frac{\left(n^2+n\right)^2-\left(n^4+1\right)}{\scriptsize\left(n^2+n\right)^{3/2}+\left(n^2+n\right)\left(n^4+1\right)^{1/4}+\left(n^2+n\right)^{1/2}\left(n^4+1\right)^{1/2}+\left(n^4+1\right)^{3/4}}\\
&=\frac{2+\frac1n-\frac1{n^3}}{\scriptsize\left(1+\frac1n\right)^{3/2}+\left(1+\frac1n\right)\left(1+\frac1{n^4}\right)^{1/4}+\left(1+\frac1n\right)^{1/2}\left(1+\frac1{n^4}\right)^{1/2}+\left(1+\frac1{n^4}\right)^{3/4}}
\end{align}
$$
</s>