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L'Hôpital's Rule

Unit 3: More with Derivatives

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L'Hôpital's Rule

Examples written by:
Kalyan Karamsetty
Image Credit:
Rob Musson

Examples Coming Soon!

Reference

The following are the seven indeterminate forms. \begin{array} {|r|r|}\hline \frac{0}{0} & \frac{\infty}{\infty} & 0\cdot\infty & \infty-\infty & 0^{0} & 1^{\infty} & \infty^{0} \\ \hline \end{array} \[\] If \(\mathop {\lim }\limits_{x \to a} \frac{f\left(x\right)}{g\left(x\right)}\) is equal to either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \( \mathop {\lim }\limits_{x \to a} \frac{f\left(x\right)}{g\left(x\right)}=\mathop {\lim }\limits_{x \to a} \frac{f'\left(x\right)}{g'\left(x\right)}\). This is known as L'Hôpital's rule. If the limit is equal to one of the other indeterminate forms, you must algebraically manipulate so that the limit yields either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). \[\] If the limit is equal to \(\infty-\infty\), the indeterminate difference, then we can algebraically manipulate by finding a common denominator, rationalizing it, or factoring out a common factor. If we do one of the three, the limit should then yield either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) and we will be able to use L'Hôpital's rule. \[\] If the limit is equal to \(0\cdot \pm\infty\), the indeterminate product, then we can algebraically manipulate the function by using the following formulas. \[\mathop {\lim }\limits_{x \to a} f\left(x\right)g\left(x\right) =\mathop {\lim }\limits_{x \to a} \frac{f\left(x\right)}{\frac{1}{g\left(x\right)}} \] \[\mbox{or}\] \[\mathop {\lim }\limits_{x \to a} f\left(x\right)g\left(x\right)=\mathop {\lim }\limits_{x \to a} \frac{g\left(x\right)}{\frac{1}{f\left(x\right)}}\] Doing so will yield either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), which in turn will allow us to use L'Hôpital's rule. \[\] If the limit is equal to \(0^{0}\), \(\infty^{0}\), or, \(1^{\infty}\) (the indeterminate powers) and the function is in the form \(y=f\left(x\right)^{g\left(x\right)}\), we can algebraically manipulate in the following way so that the limit yields either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) . \[y=f\left(x\right)^{g\left(x\right)}\] \[\ln\left(y\right)=\ln\left(f\left(x\right)^{g\left(x\right)}\right)=g\left(x\right)\ln\left(f\left(x\right)\right)\] \[e^{\ln\left(y\right)}=e^{g\left(x\right)\ln\left(f\left(x\right)\right)}\] \[y=e^{g\left(x\right)\ln\left(f\left(x\right)\right)}\] L'Hôpital's rule can only be used with the seven indeterminate forms.

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EX 1

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to 2}\frac{\sin\left(\pi x\right)}{x^{2}-5x+6}\]

EX 2

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to 2} \frac{2x^{3}+2x^{2}\ +2}{8x^{2}+9x^{3}+4x}\]

EX 3

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to 5} \frac{\sqrt{x+20}-5}{x-5}\]

EX 4

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to 1^+} \frac{3}{x-2}-\frac{2}{\ln\left(x-1\right)}\]

EX 5

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to \infty}\left(e^{x}+x\right)^{\frac{1}{x}}\]

EX 6

Evaluate the following limit.

\[\mathop {\lim }\limits_{x \to 0^+}\left(\tan\left(3x\right)\right)^{x}\]

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