Calculus

Limits and Continuity of Functions

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L’Hopital’s Rule

L'Hopital's Rule provides a method for evaluating indeterminate forms of type 0/0 or ∞/∞.

Let a be either a finite number or infinity.

This rule appeared in 1696 (!) in the first book on differential calculus published by French mathematician Guillaume de l'Hopital (1661−1704).

French mathematician Guillaume de l'Hopital
Fig.1 Guillaume de l'Hopital (1661-1704)

We can apply L'Hopital's rule to indeterminate forms of type \(0 \cdot \infty,\) \(\infty - \infty,\) \(0^0,\) \(1^{\infty},\) \(\infty^0\) as well. The first two indeterminate forms \(0 \cdot \infty\) and \(\infty - \infty\) can be reduced to the forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) using algebraic transformations. The indeterminate forms \(0^0,\) \(1^{\infty}\) and \(\infty^0\) can be reduced to the form \(0 \cdot \infty\) with the help of identity

\[f{\left( x \right)^{g\left( x \right)}} = {e^{g\left( x \right)\ln f\left( x \right)}}.\]

L'Hopital's rule also holds for one-sided limits.

Solved Problems

Example 1.

Find the limit \[\lim\limits_{x \to 0} {\frac{{{a^x} - 1}}{x}}.\]

Solution.

Differentiating the numerator and denominator separately, we get

\[\lim\limits_{x \to 0} \frac{{{a^x} - 1}}{x} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 0} \frac{{{{\left( {{a^x} - 1} \right)}^\prime }}}{{{{\left( x \right)}^\prime }}} = \lim\limits_{x \to 0} \frac{{{a^x}\ln a}}{1} = \ln a\lim\limits_{x \to 0} {a^x} = \ln a \cdot 1 = \ln a.\]

Example 2.

Find the limit \[\lim\limits_{x \to 2} {\frac{{\sqrt {7 + x} - 3}}{{x - 2}}}.\]

Solution.

Because direct substitution leads to an indeterminate form \(\frac{0}{0},\) we can use L'Hopital's rule:

\[\lim\limits_{x \to 2} \frac{{\sqrt {7 + x} - 3}}{{x - 2}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 2} \frac{{{{\left( {\sqrt {7 + x} - 3} \right)}^\prime }}}{{{{\left( {x - 2} \right)}^\prime }}} = \lim\limits_{x \to 2} \frac{{\frac{1}{{2\sqrt {7 + x} }}}}{1} = \frac{1}{2}\lim\limits_{x \to 2} \frac{1}{{\sqrt {7 + x} }} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6}.\]

Example 3.

Calculate the limit \[\lim\limits_{x \to 2} \left( {{\frac{4}{{{x^2} - 4}}} - {\frac{1}{{x - 2}}}} \right).\]

Solution.

Here we deal with an inderminate form of type \(\infty - \infty.\) After simple transformations, we have

\[\lim\limits_{x \to 2} \left( {\frac{4}{{{x^2} - 4}} - \frac{1}{{x - 2}}} \right) = \lim\limits_{x \to 2} \frac{{4 - \left( {x + 2} \right)}}{{{x^2} - 4}} = \lim\limits_{x \to 2} \frac{{2 - x}}{{{x^2} - 4}} = \left[ {\frac{0}{0}} \right] = \lim\limits_{x \to 2} \frac{{{{\left( {2 - x} \right)}^\prime }}}{{{{\left( {{x^2} - 4} \right)}^\prime }}} = \lim\limits_{x \to 2} \left( {\frac{{ - 1}}{{2x}}} \right) = - \frac{1}{4}.\]

Example 4.

Find the limit \[\lim\limits_{x \to \infty } {\frac{{{x^2}}}{{{2^x}}}}.\]

Solution.

Using L'Hopital's rule, we can write

\[\lim\limits_{x \to \infty } \frac{{{x^2}}}{{{2^x}}} = \left[ {\frac{\infty }{\infty }} \right] = \lim\limits_{x \to \infty } \frac{{{{\left( {{x^2}} \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^\prime }}} = \lim\limits_{x \to \infty } \frac{{2x}}{{{2^x}\ln 2}} = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{x}{{{2^x}}} = \left[ {\frac{\infty }{\infty }} \right] = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{{{{\left( x \right)}^\prime }}}{{{{\left( {{2^x}} \right)}^\prime }}} = \frac{2}{{\ln 2}}\lim\limits_{x \to \infty } \frac{1}{{{2^x}\ln 2}} = \frac{2}{{{{\left( {\ln 2} \right)}^2}}}\lim\limits_{x \to \infty } \frac{1}{{{2^x}}} = \frac{2}{{{{\left( {\ln 2} \right)}^2}}} \cdot 0 = 0.\]

See more problems on Page 2.

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