Limits of a bounded function...

Question

I actually encountered this one problem in my textbook and I had a doubt regarding the solution to this problem.

The question is as follows

And here is the solution to this problem:

Now I actually understood the first three lines of the solution but in 4th line, I didn't get what they did. In the third line, they wrote the expressions in the differential form and just canceled out the differentials in the 4th line. I don't think it is right. I have never seen doing such a thing anywhere else. So can someone validate that is this method right or wrong(with reasons)? And if this isn't right then can someone provide me with the right solution to this question?

Thank you in advance

Answer 1

This "cancellation" of the differential operators $\frac{d}{dx}$ is actually an application of L'Hôpital's rule to the limit

$$\lim_{x\to\infty}\frac{e^{-\lambda x}f(x)}{-\frac{1}{\lambda}e^{-\lambda x}}$$

Notice that the numerator and the denominator of $\frac{e^{-\lambda x}f(x)}{-\frac{1}{\lambda}e^{-\lambda x}}$ both tend to $0$ as $x\to\infty$. The numerator's limit follows from the product rule applied to $\lim_{x\to\infty}\left[e^{-\lambda x}f(x)\right]$:

$$\lim_{x\to\infty}\left[e^{-\lambda x}f(x)\right]=0\cdot L_2=0$$

By rewriting $e^{-\lambda x}f'(x)-\lambda e^{-\lambda x}f(x)$ as $\frac{d}{dx}\left[e^{-\lambda x}f(x)\right]$ and $e^{-\lambda x}$ as $\frac{d}{dx}\left(-\frac{1}{\lambda}e^{-\lambda x}\right)$, they showed that the limit of the ratio of derivatives exists and equals $L_1$:

$$\lim_{x\to\infty}\frac{\frac{d}{dx}\left[e^{-\lambda x}f(x)\right]}{\frac{d}{dx}\left(-\frac{1}{\lambda}e^{-\lambda x}\right)}=L_1$$

Since the numerator and denomiator both tend to $0$, L'Hôpital's rule implies that the limit

$$\lim_{x\to\infty}\frac{e^{-\lambda x}f(x)}{-\frac{1}{\lambda}e^{-\lambda x}}$$

exists, and equals $L_1$.