This limit does not exist as it approaches 1 from both sides.

[tmt1]

If I have this limit:

$$\lim_{{R}\to{1}} \frac{1}{R - 1}$$

I try to apply L'hopital's rule:

The derivative of 1 is 0, and the derivative of $R - 1$ is 1.

So I get $\frac{0}{1}$ which is 0. But apparently the answer is infinity.

What am I doing wrong?

 

[Greg]

It's not an indeterminate form; e.g. $\dfrac00$, $\dfrac{\infty}{\infty}$. Google "indeterminate form" for more information.

 

[MarkFL]

If I have this limit:

$$\lim_{{R}\to{1}} \frac{1}{R - 1}$$

I try to apply L'hopital's rule:

The derivative of 1 is 0, and the derivative of $R - 1$ is 1.

So I get $\frac{0}{1}$ which is 0. But apparently the answer is infinity.

What am I doing wrong?

I would actually give DNE for an answer, since:

\(\displaystyle \lim_{R\to1^{-}}\frac{1}{R-1}\ne\lim_{R\to1^{+}}\frac{1}{R-1}\)