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 $h(x) = f(x)-g(x)$
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> and $h(-1) = 0$ and $h(-2) = 2\;,$ Then $h(4)=\;,$ Where $a\_{1},a\_{2}\neq 0$
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$\bf{My\; Try::}$ 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: 
Hint let $l=n^{2/n} $ so $\log (l)=\frac {2}{n}\log (n) $ now we know $n $ grows faster than $log (n) $ so $log (l)=0$ thus $l=e^0=1$ thus limit is $1$
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,
$$=e^{\lim\_{n\to \infty}\large \frac{1/n}{1}}=e^{\lim\_{n\to \infty}\large \frac1n}=1$$