Limit involving complicated integral $$\lim_{x\to\infty} \sqrt{x} \int_0^\frac{\pi}{4} e^{x(\cos t-1)}\cos t\ dt$$
I attempted to work out the integral part, but it did not work well because of the existence of the e part. So whether there is a better and more convient way to calculate this limit.
Thanks!!
 A: Laplace's method provides a relatively simple result.  Here, one sees that the integral is dominated by the contribution in a small neighborhood about $t=0$.  In this neighborhood, $1-\cos{t} \sim t^2/2$.  The neighborhood is thus defined by $0 \lt x t^2/2 \lt \epsilon \implies 0 \lt t \lt \sqrt{2 \epsilon/x}$.  Because the integral contributions outside this neighborhood are exponentially small, we may simply approximate the integral with
$$\int_0^{\infty} dt \, e^{-x t^2/2} = \sqrt{\frac{\pi}{2 x}}$$
Note that we are OK with replacing the cosine outside the exponential by $1$ because, to this order, it doesn't contribute.  The limit you seek is thus $\sqrt{\pi/2}$.
ADDENDUM
Just for laughs, let's verify numerically that this is in fact the correct leading asymptotic behavior.  Here are a few Mathematica commands and output:

Integral to be evaluated. 
$$f(\text{x$\_$})\text{:=}\text{NIntegrate}\left[\cos (t) \exp (-x (1-\cos (t))),\left\{t,0,\frac{\pi }{4}\right\}\right]$$
Log-Log plot of the integral superimposed with its leading asymptotic behavior:
$$\text{Plot}\left[\left\{\log _{10}\left(f\left(10^x\right)\right),\log _{10}\left(\sqrt{\frac{\pi }{2}}\right)-\frac{x}{2}\right\},\{x,0,4\}\right]$$


Log-Log plot of difference between integral and leading asymptotic behavior.  Note the slope is $-3/2$, the exponent of the next order behavior:
$$\text{Plot}\left[\log _{10}\left(\left| f\left(10^x\right)-\sqrt{\frac{\pi }{2\ 10^x}}\right| \right),\{x,2,4\}\right]$$

A: I agree with the argument of Ron Gordon, which provides a very concise proof.
Another way : Consider the same integral, but with the upper boundary $= pi/2$ instead of $pi/4$. The limit is the same (same argument already given by Ron Gordon). The advantage is that a closed form is known for the integral. Then, the limit for $x$ can be directly derived :
$$\sqrt x\int_0^{\pi/2}e^{x(\cos (t)-1)}\cos(t)dt=\sqrt x\dfrac\pi2e^{-x}\big(I_1(x)+L_{-1}(x)\big).$$
$I_1(x)$ is the modified Bessel function of the first kind.  $L_{-1}(x)$ is the modified Struve function.  For $x\rightarrow\infty:$ the functions $I_1(x)$ and $L_{-1}(x)$ are both equivalent to: $$e^x\left(\dfrac1{\sqrt{2\pi x}}+O\left(\dfrac1{x^{3/2}}\right)\right).$$ As a consequence: $$\sqrt x\int_0^{\pi/2}e^{x(\cos (t)-1)}\cos(t)dt
\approx \sqrt x \dfrac\pi2e^{-x}\left(2e^x\left(\dfrac1{\sqrt{2\pi x}}+O\left(\dfrac1{x^{3/2}}\right)\right)\right) \\ \ \\
\lim\limits_{x\to\infty}\sqrt x\int_0^{\pi/2}e^{x(\cos (t)-1)}\cos(t)dt=\sqrt{\dfrac\pi2}.$$
Corrected : There was a mistake in the calculus of the asymptotic series.
