Recipe

Evaluate a limit with L'Hopital's Rule

Use when: \lim_{x\to c}\dfrac{f(x)}{g(x)} comes out to a genuine 0/0 or \infty/\infty form

Goal: the limit's value, reached by differentiating the numerator and denominator separately rather than the original expression

Taught in Indeterminate forms and L'Hopital's Rule

1. 1

   Plug in x=c. Only continue if the result is exactly 0/0 or \infty/\infty.

   • the result is any other number — that number IS the limit; L'Hopital's Rule does not apply and would give a wrong answer if used anyway.
   • the result is 0/0 or \infty/\infty — continue to Step 2.

   Why: the rule is only proven to work for these two forms. Applying it elsewhere doesn't just fail to help, it can produce a confidently wrong number.

2. 2

   Compute f'(x) and g'(x) on their own — this is NOT the quotient rule, since nothing is being divided yet.

   Example

   For \dfrac{e^{2x}-1}{x}: f'(x)=2e^{2x} and g'(x)=1.

   Why: L'Hopital's Rule replaces the original ratio with the ratio of derivatives. Mixing in the quotient rule instead would compute something else entirely.

3. 3

   Take \lim_{x\to c}\dfrac{f'(x)}{g'(x)}.

   Example

   \lim_{x\to0}\dfrac{2e^{2x}}{1}=2e^0=2.

   • the new limit evaluates to a number — that number is the original limit's value too.
   • the new ratio is STILL 0/0 or \infty/\infty — apply the rule again, differentiating f' and g' in turn.

   Why: the rule guarantees the two limits are equal whenever the original was truly indeterminate, so evaluating the simpler one answers the original question.