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Lopital Rule | L’Hôpital’s Rule

If both functions

f(x), g(x)

Are defined in an open interval containing point c and have a derivative in the interval, except maybe at c. And fulfill the following conditions:

  1. Both functions are tending to zero or both are tending to plus / minus infinity separately, that is we are getting the indeterminate forms:

\frac{0}{0}, \frac{\pm \infty}{\pm \infty}

2. In the interval the following holds:

g'(x)\neq 0

3. And the following holds

\lim_{x \rightarrow c} \frac{f'(x)}{g'(x)}=L

Then the following holds

\lim _ { x \rightarrow c} \frac{f(x)}{g(x)}=\lim _ { x \rightarrow a} \frac{f'(x)}{g'(x)}=L

Notes:

  1. The point c can be infinity.
  2. Notice that the we derive the numerator and denominator separately. We do NOT use the quotient rule.

How to use Lopital Rule in other indeterminate forms

  1. The following indeterminate form

0\cdot (\pm \infty)

Can be transformed to an indeterminate form that is good for Lopital Rule in this way

0\cdot \infty = \frac{0}{\frac{1}{\infty}}=\frac{0}{0}

Or

0\cdot \infty = \frac{\infty}{\frac{1}{0}}=\frac{\infty}{\infty}

2. When getting the following indetrminate form

\infty - \infty

It is worth checking if calculating a common denominator will give one fraction that provides the indeterminate forms of Lupital Rule.

3. When getting the following indeterminate forms

1^{\infty}, 0^0, {\infty}^0

One can use the formula

x=e^{\ln x}

In order to get

{f(x)}^{g(x)}=e^{\ln {f(x)}^{g(x)}}=

=e^{g(x)\ln f(x)}=

Then enter the limit to the power and get a limit of a multiplication

e^{\lim _ { x \rightarrow c} {g(x)\ln f(x)}}

Press here for exercises and solution using Lopital Rule

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