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.

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} $
