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: 
Write $n^{1/n} = 1 + a\_n.$ We have $0<a\_n$ for all $n.$ The binomial formula shows

$$n = (1 + a\_n)^n = 1 + na\_n + [n(n-1)/2]a\_n^2 + \cdots$$

for $n>1.$ Thus for $n>1$ we have $n>[n(n-1)/2]a\_n^2,$ which implies $0 < a\_n < \sqrt {2/(n-1)}.$ It follows that $a\_n \to 0,$ giving $n^{1/n} \to 1.$
Let $y = (3n^2)^{\frac 1n}$. Then $\ln y = \dfrac{\ln(3n^2)}{n}$. Note $
\dfrac{d}{dn}ln(3n^2) = \dfrac{6n}{3n^2}=\dfrac 2n$. So, by L'Hospital's rule,$\displaystyle \lim\_{n \to \infty} \ln y = \lim\_{n \to \infty}\dfrac{\ln(3n^2)}{n} = \lim\_{n \to \infty}\dfrac{2}{n^2} = 0$. Because $\ln$ is continuous, this implies that $\displaystyle \ln\left(\lim\_{n \to \infty}y\right) = 0$. Hence $\displaystyle \lim\_{n \to \infty}y = e^0 = 1.$