Calculus BC · Unit 4
Use L'Hospital's Rule on 0/0 and ∞/∞ forms, and rewrite a product or a difference first.
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Check the form before you differentiate
L'Hospital's Rule applies to the forms \(0/0\) and \(\infty/\infty\).
Differentiate the numerator and the denominator separately. Then take the limit again.
A form that is not indeterminate stops here
If the numerator approaches a nonzero number and the denominator approaches 0, the limit is infinite or does not exist. If the denominator approaches a nonzero number, substitute.
You may need the rule more than once
After one differentiation, look at the new form. If it is still \(0/0\) or \(\infty/\infty\), differentiate again.
Rewrite a product or a difference first
A form \(0\cdot\infty\) or \(\infty - \infty\) is not ready for the rule. Write it as a quotient, then check the form.
Each limit is 0/0. Differentiate once.
Substitute: \(\dfrac{\sin 0}{0}\) has the form \(0/0\).
\(\lim_{x\to 0}\dfrac{3\cos(3x)}{1}\)
\(3\cos 0\)
\(3\)
Answer: \(3\)
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{e^x}{1}\)
\(1\)
Answer: \(1\)
Substitute \(x = 2\): the form is \(0/0\).
\(\lim_{x\to 2}\dfrac{2x}{1}\)
\(4\)
Answer: \(4\)
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{\sin x}{1}\)
\(0\)
Answer: \(0\)
The form is \(0/0\).
\(\lim_{x\to 1}\dfrac{1/x}{1}\)
\(1\)
Answer: \(1\)
The form is \(\infty/\infty\).
\(\lim_{x\to\infty}\dfrac{3}{2}\)
\(\dfrac{3}{2}\)
Answer: \(\dfrac{3}{2}\)
Two rounds, a rewrite, or a limit where the rule does not apply.
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{2e^{2x} - 2}{2x}\)
That form is still \(0/0\).
\(\lim_{x\to 0}\dfrac{4e^{2x}}{2}\)
\(2e^{0}\)
\(2\)
Answer: \(2\)
The form is \(0\cdot(-\infty)\).
\(x\ln x = \dfrac{\ln x}{1/x}\)
As \(x\to 0^+\), the form is \(\infty/\infty\) in magnitude: the numerator goes to \(-\infty\) and the denominator to \(+\infty\).
\(\lim_{x\to 0^+}\dfrac{1/x}{-1/x^2}\)
\(\lim_{x\to 0^+} (-x)\)
\(0\)
Answer: \(0\)
The form is \(\infty/\infty\).
\(\lim_{x\to\infty}\dfrac{2x}{e^x}\)
Still \(\infty/\infty\).
\(\lim_{x\to\infty}\dfrac{2}{e^x}\)
\(0\)
Answer: \(0\)
The numerator approaches 0 and the denominator is 5.
The form is \(0/5\), which is not indeterminate.
\(\dfrac{0}{5} = 0\)
Answer: \(0\)
The numerator approaches 2 and the denominator approaches 0.
The form is not \(0/0\).
From the right the quotient is positive and unbounded. From the left it is negative and unbounded.
The two-sided limit does not exist.
Answer: does not exist
The form is \(\infty/\infty\).
\(\lim_{x\to\infty}\dfrac{1/x}{1}\)
\(0\)
Answer: \(0\)
Three rounds, a difference of roots, or a product written as a quotient.
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{1 - \cos x}{3x^2}\)
Still \(0/0\).
\(\lim_{x\to 0}\dfrac{\sin x}{6x}\)
\(\dfrac{1}{6}\lim_{x\to 0}\dfrac{\sin x}{x}\)
\(\dfrac{1}{6}\)
Answer: \(\dfrac{1}{6}\)
The form is \(\infty - \infty\).
\(\dfrac{x - \sin x}{x\sin x}\)
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{1 - \cos x}{\sin x + x\cos x}\)
Still \(0/0\).
\(\lim_{x\to 0}\dfrac{\sin x}{2\cos x - x\sin x}\)
\(\dfrac{0}{2}\)
\(0\)
Answer: \(0\)
The form is \(\infty\cdot 0\).
\(\dfrac{e^{1/x} - 1}{1/x}\)
As \(x\to\infty\), the form is \(0/0\).
Let \(t = 1/x\). As \(x\to\infty\), \(t\to 0\).
\(\lim_{t\to 0}\dfrac{e^t - 1}{t}\)
\(1\)
Answer: \(1\)
The form is \(\infty - \infty\).
Multiply by the conjugate.
\(\dfrac{(x^2 + 4x) - x^2}{\sqrt{x^2 + 4x} + x}\)
\(\dfrac{4x}{\sqrt{x^2 + 4x} + x}\)
For \(x > 0\), divide the numerator and the denominator by \(x\).
\(\dfrac{4}{\sqrt{1 + \dfrac{4}{x}} + 1}\)
\(\dfrac{4}{1 + 1}\)
\(2\)
Answer: \(2\)
\(\dfrac{\ln x}{x}\) has the form \(\infty/\infty\), and the next quotient is \(1/x\), which approaches 0.
The other three limits are 3, 1, and \(3/2\).
Answer: A. \(\lim_{x\to\infty}(\ln x)/x = 0\)
The form is \(0/0\).
\(\lim_{x\to 0}\dfrac{2\sin(2x)}{2x}\)
\(\lim_{x\to 0}\dfrac{\sin(2x)}{x}\)
\(\lim_{x\to 0}\dfrac{2\cos(2x)}{1}\)
\(2\)
Answer: \(2\)
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