L'Hôpital's Rule for $ \frac{\infty}{\infty-\infty} $?

Solution 1:

$${\sec x+\tan x\over\sec x-\tan x}={1+\sin x\over1-\sin x}$$

Solution 2:

Solution I

Note that the integrand is even and then you have that: $$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ but $$\int^{\infty}_{0}\frac{1}{x+1} \longrightarrow \infty\leq\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ and the improper integral diverges.

This first solution is very similar to Will Jagy's solution you may find in a message above.

Q.E.D.

Solution II

Also observe that the integrand is the derivative of $\sinh^{-1}$(x). The conclusion is evident.

Q.E.D.

Solution III

Another elementary solution?

$$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$ $$\int^{\infty}_{0}\frac{x}{x^2+1}= \lim_{x\to\infty}\frac{1}{2} \ln (x^2+1) \longrightarrow \infty\leq\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$

Q.E.D.

Solution IV

Could the inverse of the integrand allow us to evaluate the improper integral without being necessary to use any integration? (see the real positive axes)

Solution V

Consider again that $$\int^{\infty}_{-\infty}\frac{1}{\sqrt{x^{2}+1}} dx=2\int^{\infty}_{0}\frac{1}{\sqrt{x^{2}+1}}$$

then you do 2 things. Firstly, note $x = \tan y$ and for the result you get, use the nice work of Raymond Manzoni here, namely the first 3 rows of his answer and you're nicely done.

(of course, it is enough to compute the limit to $\frac{\pi}{2}$, but the approach from the link is worth to be seen)