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Mathematics LibreTexts

3.2: L'Hôpital's Rule

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    495
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    Proof of L'Hôpital's Rule

    Goal: Easily find

    \[ \lim _{x \rightarrow 0} \dfrac{\sin\, x}{x}.\nonumber \]

    Suppose that \(f\) and \(g\) are continuous functions and

    \[ f(a) = g(a) = 0.\nonumber \]

    Recall the mean value theorem states that

    \[ f'(c) = \dfrac{f(b)-f(a)}{b-a} \nonumber \]

    so that

    \[ \dfrac{ f'(c)}{g'(c)} = \dfrac{\dfrac{f(b)-f(a)}{b-a}}{\dfrac{g(b)-g(a)}{b-a}} = \dfrac{f(b)-f(a)}{g(b)-g(a)}.\nonumber \]

    Let

    \[a = 0.\nonumber \]

    Then

    \[ \dfrac{f'(c)}{g'(c)} = \dfrac{f(b)-f(0)}{g(b)-g(0)} = \dfrac{f(b)-0}{g(b)-0} =\dfrac{f(b)}{g(b)}\nonumber \]

    so that

    \[ \lim_{b \rightarrow 0} \dfrac{f(b)}{g(b)} = \lim_{c \rightarrow 0} \dfrac{f'(c)}{g'(c)}.\nonumber \]

    Hence

    \[ \lim_{x \rightarrow 0} \dfrac{\sin\,x}{x} = \lim_{x \rightarrow 0} \dfrac{\cos \, x}{1} = 1.\nonumber \]

    Definition: L'Hôpital's Rule

    Let

    \[ f(c) = g(c) = 0 \nonumber \]

    then

    \[ \lim_{x \rightarrow c} \dfrac{f(x)}{g(x)} = \lim_{x \rightarrow c} \dfrac{f'(x)}{g'(x)}. \nonumber\nonumber \]

    Example \(\PageIndex{1}\)

    \[ \lim_{x \rightarrow 0} \dfrac{e^x - 1}{x} = \lim_{x \rightarrow 0} \dfrac{e^x }{1} = 1. \nonumber\nonumber \]

    Exercise \(\PageIndex{1}\)

    Determine the following limits if they exist

    1. \( \lim_{x \rightarrow 1} \dfrac{\ln x}{x^2 -1}\)
    2. \( \lim_{x \rightarrow 0} \dfrac{\sin x}{x + 1}\)
    3. \( \lim_{x \rightarrow \infty} \dfrac{e^{-x}}{x^2}\).

    Hidden Forms of L'Hôpital's Rule

    We can also use L'Hôpital's rule when we have expressions of the form

    \( 0 \times \infty\), \( \infty^0\), and \( \infty - \infty \).

    Example \(\PageIndex{2}\) : \(0\times\infty\)

    \[\begin{align*} \lim_{x\to{\infty}}\big(\arctan x -\dfrac{\pi}{2}e^x\big) &= \lim_{x\to{\infty}} \Big(\dfrac{\arctan x -\dfrac{\pi}{2}}{e^{-x}}\Big) \\[4pt] &=\lim_{x\to{\infty}} \dfrac{\dfrac{1}{1+x^2}}{-e^{-x}} \\[4pt] &= -\lim_{x\to{\infty}}\dfrac{e^x}{1+x^2} \\[4pt] &= -\lim_{x\to{\infty}} \dfrac{e^x}{2x} \\[4pt] &= -\lim_{x\to{\infty}}\dfrac{e^x}{2} \\[4pt]&= -\infty. \end{align*}\nonumber \]

    Example \(\PageIndex{3}\): \((\infty)^0\)

    \[\begin{align*} \lim_{x\to{\infty}}\big(1+\dfrac{1}{x}\big)^x &= e^{ \lim_{x\to{\infty}}x\ln(1+\frac{1}{x})} \\[4pt] &= e^{ \lim_{x\to{\infty}} \frac{ \ln(1+\frac{1}{x})}{\frac{1}{x}}} \\[4pt] &= e^{ \lim_{x\to{\infty}} \frac{ -\frac{1}{x^2}\big(\frac{1}{1+\frac{1}{x}}\big)}{-\frac{1}{x^2}}} \\[4pt] &= e^{\lim_{x\to{\infty}} \frac{1}{1+\frac{1}{x}}} \\[4pt] &= e^1 \\[4pt] &= e. \end{align*}\nonumber \]

    Example \(\PageIndex{3}\): \(\infty - \infty\)

    \[\begin{align*} \lim_{x\to{1^+}} \Big[\dfrac{1}{x^2-1}-\dfrac{1}{\ln x}\big] &= \lim_{x\to{1^+}}\Big[ \dfrac{\ln x -(x^2-1)}{(x^2-1)\ln x} \Big] \\[4pt] &= \lim_{x\to{1^+}}\Big[ \dfrac{\dfrac{1}{x}-2x}{2x\ln x+\dfrac{x^2-1}{x}} \Big] \\[4pt] &= \dfrac{1-2}{0}. \end{align*}\nonumber \]

    Exercise \(\PageIndex{2}\)

    Evaluate

    \[ \lim_{x \rightarrow 0^+} x^x.\nonumber \]

    Larry Green (Lake Tahoe Community College)

    • Integrated by Justin Marshall.

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