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: 
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.
see my nice answer:
$$\lim\_{n\to \infty}n^{1/n}=e^{\lim\_{n\to \infty}\large \frac1n\ln n}$$ 
apply l'hopital's rule,
$$=e^{\lim\_{n\to \infty}\large \frac{1/n}{1}}=e^{\lim\_{n\to \infty}\large \frac1n}=1$$