Goal: Apply L’Hôpital’s Rule to evaluate limits of indeterminate forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\), and evaluate limits of the indeterminate forms \(0 \cdot \infty\), \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\).
1. Indeterminate Forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\)
Example 1.1: \(\displaystyle \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}}\)
Evaluate the limit using L’Hôpital’s Rule.
Solution: Direct substitution gives \(\frac{\infty}{\infty}\), so we can apply L’Hôpital’s Rule.
Therefore, \(\displaystyle \lim_{x \to \infty} \frac{\ln x}{\sqrt{x}} = 0\).
Example 1.2: \(\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{3x^2}\)
Evaluate the limit using L’Hôpital’s Rule.
Solution: Direct substitution gives \(\frac{0}{0}\), so we can apply L’Hôpital’s Rule.
Therefore, \(\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{3x^2} = \frac{1}{6}\).
Example 1.3: \(\displaystyle \lim_{x \to 0} \frac{e^{7x} - \cos(2x)}{\tan(3x)}\)
Evaluate the limit using L’Hôpital’s Rule.
Solution: Direct substitution gives \(\frac{0}{0}\), so we can apply L’Hôpital’s Rule.
Therefore, \(\displaystyle \lim_{x \to 0} \frac{e^{7x} - \cos(2x)}{\tan(3x)} = \frac{7}{3}\).
Example 1.4: \(\displaystyle \lim_{x \to 0} \frac{\sin x}{1 - x^2}\)
Evaluate the limit. Note: Do not apply L’Hôpital’s Rule blindly; consider direct substitution and simplification.
Solution: Direct substitution gives:
Answer: \(\displaystyle \lim_{x \to 0} \frac{\sin x}{1 - x^2} = 0\)
2. Indeterminate Form \(0 \cdot \infty\)
Example 2.1: \(\displaystyle \lim_{x \to 0^+} x \cdot \ln(2x)\)
Evaluate the limit using L’Hôpital’s Rule. Note that this is an indeterminate form \(0 \cdot (-\infty)\).
Solution: Rewrite the product as a quotient to apply L’Hôpital’s Rule:
Now as \(x \to 0^+\), the limit is of the form \(\frac{-\infty}{\infty}\), which is suitable for L’Hôpital’s Rule.
Therefore, \(\displaystyle \lim_{x \to 0^+} x \cdot \ln(2x) = 0\).
Example 2.2: \(\displaystyle \lim_{x \to \infty} 2x \cdot \tan\left(\frac{1}{3x}\right)\)
Evaluate the limit using L’Hôpital’s Rule. Note this is an indeterminate form \( \infty \cdot 0 \).
Solution: Rewrite the product as a quotient to apply L’Hôpital’s Rule:
As \(x \to \infty\), this becomes \(\frac{0}{0}\), which allows L’Hôpital’s Rule.
Therefore, \(\displaystyle \lim_{x \to \infty} 2x \cdot \tan\left(\frac{1}{3x}\right) = \frac{2}{3}\).
Remark: Consider the standard limit for small angles:
This is an indeterminate form \(\frac{0}{0}\), so we can apply L’Hôpital’s Rule:
Hence, \(\displaystyle \lim_{x \to 0} \frac{\tan x}{x} = 1\).
This result is useful when handling limits involving \(\tan\) as \(x \to 0\), like in Example 2.2.
3. Indeterminate Form \(\infty - \infty\)
Example 3.1: \(\displaystyle \lim_{x \to \infty} \left(\sqrt{x^2 - 5x + 7} - x\right)\)
Evaluate the limit. This is an indeterminate form \(\infty - \infty\).
Solution: When you have \(\infty - \infty\), a common technique is to rationalize using the conjugate:
Factor \(x\) from numerator and denominator to simplify the limit:
Now take the limit as \(x \to \infty\):
Alternative Method:
Strategy: Instead of rationalizing, factor \(x\) from the square root to rewrite the limit as a product that gives \(\infty \cdot 0\).
Now as \(x \to \infty\), we see it’s an indeterminate form \(\infty \cdot 0\). Rewrite as a quotient for L’Hôpital:
Now apply L’Hôpital’s Rule to the \(\frac{0}{0}\) form:
Compute derivatives carefully and then take the limit. You will get the same answer:
4. Indeterminate Forms \(1^\infty, 0^0, \infty^0\)
Example 4.1: \(\displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{7x}\)
Evaluate the limit. This is an indeterminate form \(1^\infty\).
Solution: For \(1^\infty\) forms, take the natural logarithm first:
Now consider the limit of \(\ln y\) as \(x \to \infty\):
This is an indeterminate form \(\infty \cdot 0\). Rewrite it as a quotient for L’Hôpital’s Rule:
Apply L’Hôpital’s Rule to the \(\frac{0}{0}\) form:
Therefore, \(\ln y \to 21\) as \(x \to \infty\), so
Answer: \(\displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{7x} = e^{21}\)
5. Summary
| # | Type | Example |
|---|---|---|
| 1 | L'Hôpital's Rule (\(0/0\) or \(\infty/\infty\)) |
\(\displaystyle \lim_{x\to0} \frac{\sin x}{x} = \frac{\cos 0}{1} = 1\) \(\displaystyle \lim_{x\to\infty} \frac{\ln x}{x} = \lim_{x\to\infty} \frac{1/x}{1} = 0\) \(\displaystyle \lim_{x\to0} \frac{e^{2x}-1}{x} = \lim_{x\to0} \frac{2 e^{2x}}{1} = 2\) \(\displaystyle \lim_{x \to 0} \frac{1 - \cos x}{x} = \lim_{x \to 0} \frac{(\,1 - \cos x\,)'}{(x)'} = \lim_{x \to 0} \frac{\sin x}{1} = 0\) |
| 2 | Indeterminate form \(\infty \cdot 0\) | \(\displaystyle \lim_{x\to 0} x \ln x = \lim_{x\to 0} \frac{\ln x}{1/x} = \lim_{x\to 0} \frac{1/x}{-1/x^2} = \lim_{x\to 0} -x = 0\) |
| 3 | Indeterminate exponential \(1^\infty, 0^0, \infty^0\) |
\(\displaystyle \lim_{x\to 0} (1+x)^{1/x}\) Let \(\displaystyle y = (1+x)^{1/x}\), then \(\displaystyle \ln y = \frac{\ln(1+x)}{x} \Rightarrow \lim_{x\to 0} \ln y = 1 \Rightarrow \lim_{x\to 0} y = e\) \(\displaystyle \lim_{x\to\infty} \left(1+\frac{1}{x}\right)^x\) Let \(\displaystyle y = \left(1+\frac{1}{x}\right)^x\), then \(\displaystyle \ln y = x \ln(1+1/x) \Rightarrow \lim_{x\to\infty} \ln y = 1 \Rightarrow \lim_{x\to\infty} y = e\) \(\displaystyle \lim_{x\to\infty} x^{1/x}\) Let \(\displaystyle y = x^{1/x}\), then \(\displaystyle \ln y = \frac{\ln x}{x} \Rightarrow \lim_{x\to\infty} \ln y = 0 \Rightarrow \lim_{x\to\infty} y = e^0 = 1\) |
| 4 | Direct substitution | \(\displaystyle \lim_{x\to2} (x^2+3) = 2^2 + 3 = 7\) |
| 5 | Indeterminate \(0/0\) (factor / simplify) | \(\displaystyle \lim_{x\to2} \frac{x^2-4}{x-2} = \lim_{x\to2} \frac{(x-2)(x+2)}{x-2} = \lim_{x\to2} (x+2) = 4\) |
| 6 | Limit at infinity (rational) | \(\displaystyle \lim_{x\to\infty} \frac{3x^2+1}{2x^2+5x} = \lim_{x\to\infty} \frac{x^2(3+1/x^2)}{x^2(2+5/x)} = \frac{3}{2}\) |
| 7 | Root / conjugate | \(\displaystyle \lim_{x\to4} \frac{\sqrt{x}-2}{x-4} = \lim_{x\to4} \frac{(\sqrt{x}-2)(\sqrt{x}+2)}{(x-4)(\sqrt{x}+2)} = \lim_{x\to4} \frac{x-4}{(x-4)(\sqrt{x}+2)} = \frac{1}{4}\) |