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.
 A: One simple method is to try adding terms to give upper and lower bounds for the given roots. One may confirm (and not hard to find, either)
$$ \left( n + \frac{1}{2} - \frac{1}{8n} \right)^2 < n^2 + n < \left( n + \frac{1}{2}  \right)^2 $$
$$ n^4 < n^4 + 1 < \left( n + \frac{1}{4n^3}   \right)^4 $$
Together, we get
$$ \frac{1}{2} -  \frac{1}{8n} - \frac{1}{4n^3} \; \;  < \; \; \sqrt{n^2+n}-\sqrt[4]{n^4+1} \; \; < \; \; \frac{1}{2}  $$
Some people call this the Squeeze Theorem, the limit is $\frac{1}{2} $
A: 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,
$$
\begin{align}
\sqrt{n^2+n}-\sqrt[4]{n^4+1}
&=\frac{\left(n^2+n\right)^2-\left(n^4+1\right)}{\scriptsize\left(n^2+n\right)^{3/2}+\left(n^2+n\right)\left(n^4+1\right)^{1/4}+\left(n^2+n\right)^{1/2}\left(n^4+1\right)^{1/2}+\left(n^4+1\right)^{3/4}}\\
&=\frac{2+\frac1n-\frac1{n^3}}{\scriptsize\left(1+\frac1n\right)^{3/2}+\left(1+\frac1n\right)\left(1+\frac1{n^4}\right)^{1/4}+\left(1+\frac1n\right)^{1/2}\left(1+\frac1{n^4}\right)^{1/2}+\left(1+\frac1{n^4}\right)^{3/4}}
\end{align}
$$
A: it must be $$\frac{(n^2+n)^2-(n^4+1)}{(\sqrt{n^2+n}+\sqrt[4]{n^4+1})(n^2+n+\sqrt{n^4+1})}$$
A: Let $t = \frac1n$ 
Then use the derivative formula at $t = 0$ , 
$$ \begin{align}\lim_{n\to\infty}\sqrt{n^2+n}-\sqrt[4]{n^4+1} &= \lim_{n\to\infty} n\left(\sqrt{\frac{1}{n}+1}-\sqrt[4]{\frac{1}{n^4}+1}\right)  \\&= \lim_{t\to0}\frac{1}t(\sqrt{t+1}-\sqrt[4]{t^4+1})\\&= \lim_{t\to 0}\frac{\sqrt{t+1}-1}{t} - \frac{\sqrt[4]{t^4+1}-1}{t} \\&=\left(\sqrt{t+1}\right)'\bigg|_{t =0 } -\left(\sqrt[4]{t^4+1}\right)'\bigg|_{t =0 }= \frac{1}{2}\end{align} $$
