$\lim_{\epsilon \rightarrow 0} \frac 1{\epsilon} \int_0 ^{\infty} f(x ,\epsilon) dx$

Check if the following limit exist and find the value if it exist .

$\lim_{\epsilon \rightarrow 0} \frac 1{\epsilon}\int_0 ^{\infty} e^{-x/{\epsilon} }(\cos (3x)+ x^2 +\sqrt{x+4} ) dx $

My Attempt :-

For $ X > 0 $

Let $I( X ,\epsilon)$

$= \int_0 ^X - \frac d{dx} (e^{-x/{\epsilon} })(\cos (3x)+ x^2 +\sqrt{x+4} ) dx $ .

Integrating by parts ,

$=-\big[ e^{-x/{\epsilon} }(\cos (3x)+ x^2 +\sqrt{x+4}) \big]_0 ^X + \int_0 ^X e^{-x/{\epsilon} } \big( -3\sin (3x) +2x + \frac 1{2\sqrt{x+4}} \big) dx $

$\big[ 3- e^{-X/{\epsilon} }(\cos (3X)+ X^2 +\sqrt{X+4})\big]+\int_0 ^X e^{-x/{\epsilon} } \big( -3\sin (3x) +2x + \frac 1{2\sqrt{x+4}} \big) dx $

Let $\epsilon >0$ , Then

$\lim_{X \rightarrow \infty} I(x ,\epsilon)=3+ \lim_{X \rightarrow \infty} \int_0 ^X e^{-x/{\epsilon} } \big( -3\sin (3x) +2x + \frac 1{2\sqrt{x+4}} \big) dx $

Again $\int _0 ^X e^{-x/{\epsilon} }2x dx $

$=-2\epsilon X e^{-X/{\epsilon} } +2{\epsilon}^2(1-e^{-X/{\epsilon} }) $

Taking $\lim_{X \rightarrow \infty}\int _0 ^X e^{-x/{\epsilon} }2x dx=2{\epsilon}^2 $

Again $ \int e^{ax}\sin (bx)= \frac{e^{ax}}{a^2+b^2} (a\sin(bx) -b\cos(bx))$

So $ \int_0 ^X e^{-x/{\epsilon}}\sin (3x) =-\frac{\epsilon e^{-X/{\epsilon}}}{1+3{\epsilon}^2}(\sin (3X)+ 3\epsilon \cos(3X))+\frac{3{\epsilon}^2}{1+3{\epsilon}^2} $

Taking limit as $X\rightarrow \infty$ , the above integral converges to $\frac{3{\epsilon}^2}{1+3{\epsilon}^2} $

The middle integrand $\frac{e^{-x/{\epsilon}}}{\sqrt{x+4}}< \frac{e^{-x/{\epsilon}}}{2} $

And so integrating and taking limit as $X\rightarrow \infty$ , we get

$\lim_{X\rightarrow \infty} \int_0 ^X \frac{e^{-x/{\epsilon}}}{\sqrt{x+4}} <\frac{\epsilon}2 $

Is there a problem with the above boundedness?

The above limits are all for $\epsilon >0$. How to tackle the limits $\epsilon <0$ ?

The answer is given $3$