Evaluate the following limit: $\lim_{k \to \infty} \int_0^1 e^{- (k^2x^2/2)} dx$ $\mathbf {The \ Problem \ is}:$
  Find   $$\lim_{k \to \infty} \int_0^1 e^{- (k^2x^2/2)} dx$$ 
Actually, I have been expanding  $e^{- (k^2x^2/2)}$ to find out the limit, but I can't approach further. I think it's divergent .
For any help, thanks in advance !!!
 A: You want $$\lim_{k\to\infty}\frac{1}{k}\int_0^k e^{-y^2/2}dy=0$$by squeezing viz. $$0\le\frac{1}{k}\int_0^k e^{-y^2/2}dy\le\frac{1}{k}\int_0^\infty e^{-y^2/2}dy=\frac{\sqrt{\pi/2}}{k}.$$
A: You can majorise the integrand on the interval $(0,1)$ by a constant function $\ge 1$ and apply the dominated convergence theorem. Thus, you can pass the limit under the integration. This yields $\int_0^1 \lim_{k \to \infty} e^{-\frac{k^2 x^2}{2}} \mathrm{d}x =0$.
A: The sequence $$ f_k= \left(\frac{1}{e^{x^2/2}}\right)^{k^2}$$ is bounded by $1$ on the interval $[0,1]$ and $f_k\rightarrow 0$ uniformly since $1/e^{x^2/2}$ is decreasing so it is minimum is $1/\sqrt{e}$ so 
$$\lim \int \frac{1}{e^{\frac{x^2k^2}{2}}}=\int \lim \frac{1}{e^{\frac{x^2k^2}{2}}}=0$$
A: Replacing $k^2/2$ with $t$ we can see that the desired limit is equal to $$\lim_{t\to\infty} \int_{0}^{1}e^{-tx^2}\,dx$$ The integrand is positive and clearly less than $1/(1+tx^2)$ and hence the integral is less than $$\frac{1}{\sqrt{t}}\arctan\sqrt{t} $$ The above expression tends to $0$ as $t\to\infty $ and hence by Squeeze theorem the desired limit is $0$.
