L'Hôpital's Rule/Examples/x^2 - 4 over 2 x - 4
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Example of Use of L'Hôpital's Rule
Let $f: \R \to \R$ and $g: \R \to \R$ be real functions defined as:
| \(\ds \forall x \in \R: \, \) | \(\ds \map f x\) | \(=\) | \(\ds x^2 - 4\) | |||||||||||
| \(\ds \map g x\) | \(=\) | \(\ds 2 x - 4\) |
Then:
- $\ds \lim_{x \mathop \to 2} \dfrac {\map f x} {\map g x} = 2$
Proof
We note that $\valueat {\dfrac {\map f x} {\map g x} } {x \mathop = 2}$ is undefined.
Hence we observe that L'Hôpital's Rule may be applicable.
We have that:
| \(\ds \map {f'} x\) | \(=\) | \(\ds \dfrac \d {\d x} x^2 - 4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2 x\) | Power Rule for Derivatives | |||||||||||
| \(\ds \map {g'} x\) | \(=\) | \(\ds 2 x - 4\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2\) | Power Rule for Derivatives |
Thus both $\ds \lim_{x \mathop \to 2} \map {f'} 2$ and $\ds \lim_{x \mathop \to 2} \map {g'} 2$ exist, and we can indeed continue:
| \(\ds \lim_{x \mathop \to 2} \dfrac {\map f x} {\map g x}\) | \(=\) | \(\ds \lim_{x \mathop \to 2} \dfrac {\map {f'} x} {\map {g'} x}\) | L'Hôpital's Rule | |||||||||||
| \(\ds \) | \(=\) | \(\ds \lim_{x \mathop \to 2} \dfrac {2 x} 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \dfrac 4 2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 2\) |
$\blacksquare$
Sources
- 1998: David Nelson: The Penguin Dictionary of Mathematics (2nd ed.) ... (previous) ... (next): L'Hôpital's rule (L'Hospital's rule, de L'Hôpital's rule)
- 2008: David Nelson: The Penguin Dictionary of Mathematics (4th ed.) ... (previous) ... (next): L'Hôpital's rule (L'Hospital's rule, de L'Hôpital's rule)