<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] Let $t = \frac1n$ 
Then use the derivative formula at $t = 0$ , 
$$ \begin{align}\lim_{n\to\infty}\sqrt{n^2+n}-\sqrt[4]{n^4+1} &= \lim_{n\to\infty} n\left(\sqrt{\frac{1}{n}+1}-\sqrt[4]{\frac{1}{n^4}+1}\right)  \\&= \lim_{t\to0}\frac{1}t(\sqrt{t+1}-\sqrt[4]{t^4+1})\\&= \lim_{t\to 0}\frac{\sqrt{t+1}-1}{t} - \frac{\sqrt[4]{t^4+1}-1}{t} \\&=\left(\sqrt{t+1}\right)'\bigg|_{t =0 } -\left(\sqrt[4]{t^4+1}\right)'\bigg|_{t =0 }= \frac{1}{2}\end{align} $$
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