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

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The limit in question, $$\lim_{{R}\to{1}} \frac{1}{R - 1}$$, approaches infinity as R approaches 1 from either side. The application of L'Hôpital's rule is incorrect here because the limit is not in an indeterminate form; it diverges instead. The left-hand limit ($R \to 1^{-}$) approaches negative infinity, while the right-hand limit ($R \to 1^{+}$) approaches positive infinity, indicating that the limit does not exist (DNE). This discrepancy highlights the importance of considering one-sided limits in such cases.
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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?
 
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It's not an indeterminate form; e.g. $\dfrac00$, $\dfrac{\infty}{\infty}$. Google "indeterminate form" for more information.
 
tmt said:
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:

$$\lim_{R\to1^{-}}\frac{1}{R-1}\ne\lim_{R\to1^{+}}\frac{1}{R-1}$$
 
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