Question: > 
> If $f(x) = a\_{1}x^2+b\_{1}x+c\_{1}$ and $g(x) = a\_{2}x^2+b\_{2}x+c\_{2}$ and $h(x) = f(x)-g(x)$
> 
> 
> and $h(-1) = 0$ and $h(-2) = 2\;,$ Then $h(4)=\;,$ Where $a\_{1},a\_{2}\neq 0$
> 
> 
> 

$\bf{My\; Try::}$ Here $h(x)$ is a difference of two quadratic polynomial

So $h(x)= = (a\_{1}-a\_{2})x^2+(b\_{1}-b\_{2})x+(c\_{1}-c\_{2}) = ax^2+bx+c$

And given $h(-1) = 0\;$ and $h(-2) = 2$

So $h(x)= a(x+1)(x-\beta),$ Put $\beta = -2$

$\displaystyle h(-2) = 2= A(\beta+1)\Rightarrow A=\frac{2}{\beta+1}$

So $$h(4) = 4A\cdot 5 \cdot (4-\beta) = 40\frac{4-\beta}{\beta+1}$$

Now how can i calculate $h(4)\;,$ Help required, Thanks

Answer: 
We can write $\sqrt[n]{3n^2} = (\sqrt{3}n)^\frac{2}{n} = e^\frac{2\ln(\sqrt{3}n)}{n}$. Note that for $n > 1$ we have $0 < \ln(n) < n$. If $n > 1$ then $\sqrt{\sqrt{3}n} > 1$ (since $\sqrt{3}$ > 1). Thus we have for $n > 1$
\begin{align}
0 < \ln(\sqrt{\sqrt{3}n}) < \sqrt{\sqrt{3}n}\\
0 < 2\ln(\sqrt{\sqrt{3}n}) < 2\sqrt{\sqrt{3}n}\\
0 < \ln(\sqrt{3}n) < 2\sqrt{\sqrt{3}n}\\
0 < \frac{\ln(\sqrt{3}n)}{n} <\frac{2\sqrt{\sqrt{3}} \sqrt{n}}{\sqrt{n} \sqrt{n}} = \frac{2\sqrt[4]{3}}{\sqrt{n}}
\end{align} 
Since $\lim\_{n \to \infty} \frac{2\sqrt[4]{3}}{\sqrt{n}} = 0$ we have by the squeeze theorem $\lim\_{n \to \infty} \frac{\ln(\sqrt{3}n)}{n} = 0$. Thus we have
\begin{equation}
\lim\_{n\to\infty} \sqrt[n]{3n^2} = \lim\_{n \to \infty} e^\frac{2\ln(\sqrt{3}n)}{n} = e^{2\lim\_{n\to\infty} \frac{\ln(\sqrt{3}n)}{n}} = e^{2(0)} = e^0 = 1
\end{equation}
Actually, we have a more general result, that is 
$$
\lim \_{n \to \infty} \sqrt[n]{n} = 1
,
$$
because the latter limit deduces 
$$
\lim \_{n \to \infty} \sqrt[n]{3n^2} = \lim \_{n \to \infty} \sqrt[n]{3} \cdot ( \sqrt[n]{n} )^2 = 1\cdot 1^2 = 1 
.
$$
Now we prove the first limit using the deffinition of limit, i.e. prove:
$$
\forall \epsilon > 0, ~
\exists N\_\epsilon > 0, ~~ \textrm{such that} ~~
\forall n > N\_\epsilon, ~~
\left| \sqrt[n]{n} - 1 \right| < \epsilon
$$
Let $\epsilon \in (0,1) $ arbitrarily, choose $N\_\epsilon$ to be large enough ($N\_\epsilon$ depends on $\epsilon$) such that 
$$
(1-\epsilon)^{N\_\epsilon} < N\_\epsilon < (1+\epsilon)^{N\_\epsilon}
$$
This is a quite simple task since $0< 1-\epsilon < 1$ and $1 + \epsilon > 1$, and once we make a correct choice of $N \_ \epsilon$, the above inequalities hold for any $n > N\_\epsilon$, thus, $ \left| \sqrt[n]{n} - 1 \right| < \epsilon $ for any $n > N\_\epsilon$. Game over.