<s>[INST] Limit of a sequence depending on a parameter The problem given by our teacher during the class goes as follows: Determine the limit of a sequence $a_n$ depending on the parameter $p$ if $a_n=\sqrt{4n^2+3n+5}-pn-1$.
My attempt of solution:
$$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{n\left(\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}-p-\frac{1}{n}\right)}=\lim{n} \cdot \lim{\left(\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}-p-\frac{1}{n}\right)}.$$
Now it's easy to see that all fractions under the second limit vanish and we are left with 
$\lim{n} \cdot \lim{(2-p)}$. $2-p$ is a constant value, and hence $\lim{(2-p)}=2-p$. The whole expression then becomes $$(2-p)\lim{n}.$$ Thus, for $p>2$ we get $-\infty$, and for $p<2$ the result is $+\infty$. The only unknown sitution happens for $p=2$, because we get an undefined "$0 \cdot \infty$". We've got to investigate this situation further. By plugging in $p=2$ to the original form and using a well-known technique, i.e. multiplying $\sqrt{4n^2+3n+5}-(2n+1)$ by $$\frac{\sqrt{4n^2+3n+5}+(2n+1)}{\sqrt{4n^2+3n+5}+(2n+1)},$$
we get $\lim{\sqrt{4n^2+3n+5}-(2n+1)}=-\frac{1}{4}$. And these are correct, according to the answer sheet in the textbook. My teacher, however, claims that the method I used is invalid. The problem is, I don't see any mistake here... Can you point it out for me?
 [/INST] Let's see: $$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{\frac{(\sqrt{4n^2+3n+5}-pn-1)(\sqrt{4n^2+3n+5}+pn+1)}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{4n^2+3n+5-(pn+1)^2}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n^2(4-p^2)+n(3-2p)+4}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n(4-p^2)+3-2p+\frac{4}{n}}{\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}+p+\frac{1}{n}}}$$ From here it's clear than the limit won't exist if $4-p^2 \ne 0$. If $p=2$, the limit exists ant it's equal to $-\frac{1}{2}$if $p=-2$ the limit does not exist because the denominator becomes zero. 
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