Before introducing L'Hospital's Rule, let's review evaluating limits and identify indeterminate forms.
Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^2 - 2x}{x - 2}\)
Solution:
First, attempt direct substitution:
\[\lim_{x \to 2} \frac{x^2 - 2x}{x - 2} = \frac{2^2 - 2(2)}{2 - 2} = \frac{0}{0}\]This gives us the indeterminate form \(\frac{0}{0}\). We can factor and simplify:
\[\lim_{x \to 2} \frac{x^2 - 2x}{x - 2} = \lim_{x \to 2} \frac{x(x - 2)}{x - 2} = \lim_{x \to 2} x = 2\]Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^2 - 2x}{(x - 2)^2}\)
Solution:
Direct substitution gives \(\frac{0}{0}\). Simplifying:
\[\lim_{x \to 2} \frac{x^2 - 2x}{(x - 2)^2} = \lim_{x \to 2} \frac{x(x - 2)}{(x - 2)^2} = \lim_{x \to 2} \frac{x}{x - 2}\]As \(x \to 2\), the numerator approaches 2 while the denominator approaches 0, so the limit does not exist (DNE).
Evaluate \(\displaystyle \lim_{x \to \infty} \frac{x^2 - 2x}{x - 2}\)
Solution:
This has the form \(\frac{\infty}{\infty}\), which is indeterminate. We can factor and simplify:
\[\lim_{x \to \infty} \frac{x^2 - 2x}{x - 2} = \lim_{x \to \infty} \frac{x(x - 2)}{x - 2} = \lim_{x \to \infty} x = \infty\]Important: The forms \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) are indeterminate forms. This means we cannot determine the limit without further analysis.
L'Hospital's Rule: If \(\displaystyle \lim_{x \to a} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then:
\[\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}\]provided the limit on the right exists or is \(\pm\infty\).
Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^2 - 2x}{x - 2}\) using L'Hospital's Rule.
Solution:
Direct substitution gives \(\frac{0}{0}\), so we can apply L'Hospital's Rule:
\[\lim_{x \to 2} \frac{x^2 - 2x}{x - 2} = \lim_{x \to 2} \frac{\frac{d}{dx}(x^2 - 2x)}{\frac{d}{dx}(x - 2)} = \lim_{x \to 2} \frac{2x - 2}{1} = 2(2) - 2 = 2\]Warning: L'Hospital's Rule can ONLY be applied when the limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Always check the form first!
Evaluate \(\displaystyle \lim_{x \to 2} \frac{x^2 - 2x}{(x - 2)^2}\)
Solution:
Although direct substitution gives \(\frac{0}{0}\), applying L'Hospital's Rule:
\[\lim_{x \to 2} \frac{x^2 - 2x}{(x - 2)^2} = \lim_{x \to 2} \frac{2x - 2}{2(x - 2)}\]This simplifies to:
\[\lim_{x \to 2} \frac{2(x - 1)}{2(x - 2)} = \lim_{x \to 2} \frac{x - 1}{x - 2} = \frac{2 - 1}{2 - 2} = \frac{1}{0}\]This limit does not exist (DNE). The function has a vertical asymptote at \(x = 2\).
Evaluate \(\displaystyle \lim_{x \to \infty} \frac{x^2 - 2x}{x - 2}\)
Solution:
This has the form \(\frac{\infty}{\infty}\), so we can apply L'Hospital's Rule:
\[\lim_{x \to \infty} \frac{x^2 - 2x}{x - 2} = \lim_{x \to \infty} \frac{2x - 2}{1} = \infty\]Consider \(\displaystyle \lim_{x \to 0} \frac{1 - x}{\cos x}\)
Solution:
Direct substitution gives:
\[\lim_{x \to 0} \frac{1 - x}{\cos x} = \frac{1 - 0}{\cos 0} = \frac{1}{1} = 1\]This is NOT an indeterminate form, so L'Hospital's Rule should NOT be applied.
Graph Description: A coordinate plane showing the function \(y = \frac{1-x}{\cos x}\) near \(x = 0\). The graph is smooth and continuous through the origin, approaching the point (0, 1). The function oscillates as it moves away from the origin due to the cosine term in the denominator. A horizontal line at \(y = 1\) intersects the curve at \(x = 0\), indicating the limit value.
Understanding the relative growth rates of different types of functions is crucial for evaluating limits involving indeterminate forms.
Key Principle: Any linear function (or polynomial) grows much faster than a logarithmic function.
Evaluate \(\displaystyle \lim_{x \to \infty} \frac{\ln x}{x - 1}\)
Solution:
This has the form \(\frac{\infty}{\infty}\), so we apply L'Hospital's Rule:
\[\lim_{x \to \infty} \frac{\ln x}{x - 1} = \lim_{x \to \infty} \frac{\frac{d}{dx}(\ln x)}{\frac{d}{dx}(x - 1)} = \lim_{x \to \infty} \frac{\frac{1}{x}}{1} = \lim_{x \to \infty} \frac{1}{x} = 0\]Graph Description: A coordinate plane showing two curves: \(y = \ln x\) (blue curve) and \(y = mx + b\) (red line, where \(m > 0\)). The logarithmic curve starts at the bottom left, passes through (1, 0), and increases slowly as it moves to the right with decreasing slope. The linear function is a straight line with constant positive slope. Initially, for small positive \(x\) values, the functions may be close, but as \(x\) increases to the right, the gap between the linear function (above) and the logarithmic function (below) widens dramatically, illustrating that the linear function grows much faster than the logarithmic function.
Evaluate \(\displaystyle \lim_{x \to \infty} \frac{\ln x}{x + 5}\)
Solution:
Since any linear function grows faster than a logarithmic function:
\[\lim_{x \to \infty} \frac{\ln x}{x + 5} = 0\]Growth Rate Theorem: Any polynomial grows faster than a logarithmic function. That is, for any polynomial \(p(x)\):
\[\lim_{x \to \infty} \frac{\ln x}{p(x)} = 0\]Evaluate \(\displaystyle \lim_{x \to \infty} \frac{\ln x}{5x^3 - 7x + 3}\)
Solution:
This cubic polynomial grows much faster than the straight line \(y = x\), which in turn grows faster than the logarithm. We can apply L'Hospital's Rule:
\[\lim_{x \to \infty} \frac{\ln x}{5x^3 - 7x + 3} = \lim_{x \to \infty} \frac{\frac{1}{x}}{15x^2 - 7}\]This simplifies to:
\[\lim_{x \to \infty} \frac{1}{x(15x^2 - 7)} = 0\]Since the denominator grows without bound while the numerator remains constant.
Key Principle: An exponential function grows much faster than any polynomial.
Evaluate \(\displaystyle \lim_{x \to \infty} \frac{x^5 - 12x^2 + 8}{e^x}\)
Solution:
This has the form \(\frac{\infty}{\infty}\). Applying L'Hospital's Rule:
\[\lim_{x \to \infty} \frac{x^5 - 12x^2 + 8}{e^x} = \lim_{x \to \infty} \frac{5x^4 - 24x}{e^x}\]Still \(\frac{\infty}{\infty}\), apply L'Hospital's Rule again:
\[\lim_{x \to \infty} \frac{5x^4 - 24x}{e^x} = \lim_{x \to \infty} \frac{20x^3 - 24}{e^x}\]Continue applying L'Hospital's Rule until the numerator becomes constant:
\[\lim_{x \to \infty} \frac{20x^3 - 24}{e^x} = \lim_{x \to \infty} \frac{60x^2}{e^x} = \lim_{x \to \infty} \frac{120x}{e^x} = \lim_{x \to \infty} \frac{120}{e^x} = 0\]Graph Description: A coordinate plane showing two curves: a polynomial function (black curve, relatively flat and close to the x-axis) and an exponential function \(y = e^x\) (blue curve). The exponential curve starts near the origin and rises steeply upward to the right, demonstrating rapid, unbounded growth. The polynomial, while also increasing, appears almost horizontal in comparison to the dramatic upward sweep of the exponential function, illustrating that exponential functions grow much faster than any polynomial.
Graph Description: A coordinate plane showing three curves representing different function types across the positive x-axis. The logarithmic function \(y = \ln x\) (blue curve) starts from below, passes through (1, 0), and increases slowly with decreasing slope. A polynomial function (green curve) starts from the bottom left and increases more rapidly than the logarithm, showing moderate upward curvature. An exponential function \(y = b^x\) where \(b > 1\) (purple curve) starts near the x-axis for small \(x\) values but then rises dramatically with increasing steepness, demonstrating the fastest growth rate. The relative positions illustrate the hierarchy: exponential functions grow fastest, polynomials grow at a moderate rate, and logarithmic functions grow slowest.
Hierarchy of Growth Rates: For \(b > 1\):
This gives us the following limit results as \(x \to \infty\):
\[\lim_{x \to \infty} \frac{\ln x}{x^n} = 0 \quad \text{and} \quad \lim_{x \to \infty} \frac{x^n}{e^x} = 0\]for any \(n > 0\).
Evaluate the following limits:
Solution:
Part 1: Polynomial over exponential
Since exponential functions grow faster than polynomials:
\[\lim_{x \to \infty} \frac{x^2}{3^x} = 0\]Part 2: This limit approaches \(\frac{1}{\text{large negative}}\) as \(x \to 0^+\), which equals 0. We can verify using L'Hospital's Rule if needed, but recognizing that \(x^{98}\) dominates \(\ln x\) near zero, the denominator becomes large positive, making the fraction approach 0.
\[\lim_{x \to 0^+} \frac{e^x}{\ln x + x^{98}} = 0\]L'Hospital's Rule directly applies to \(\frac{0}{0}\) and \(\frac{\infty}{\infty}\) forms. However, other indeterminate forms can be converted to these forms.
Important: The form \(0 \cdot \infty\) is indeterminate. The following are examples showing different possible outcomes:
Strategy for \(0 \cdot \infty\): Convert the product to a quotient by rewriting as:
\[f(x) \cdot g(x) = \frac{f(x)}{\frac{1}{g(x)}} \quad \text{or} \quad f(x) \cdot g(x) = \frac{g(x)}{\frac{1}{f(x)}}\]This converts the form to either \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), allowing L'Hospital's Rule to be applied.
Evaluate \(\displaystyle \lim_{x \to 0^+} x \ln x\)
Solution:
As \(x \to 0^+\), we have \(x \to 0\) and \(\ln x \to -\infty\), giving the form \(0 \cdot \infty\). We rewrite this as a fraction:
\[\lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}}\]This has the form \(\frac{-\infty}{\infty}\). Apply L'Hospital's Rule:
\[\lim_{x \to 0^+} \frac{\ln x}{\frac{1}{x}} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\frac{1}{x^2}} = \lim_{x \to 0^+} \frac{x^2}{-x} = \lim_{x \to 0^+} (-x) = 0\]Graph Description: A coordinate plane showing the function \(y = x \ln x\) for \(x > 0\). The curve starts from the positive x-axis (near \(x = 0\)) and initially descends below the x-axis into negative y-values, reaches a minimum point, then curves upward crossing the x-axis and continuing to increase for larger x-values. As \(x\) approaches 0 from the right, the function approaches 0, illustrating that the limit is 0. The graph shows that despite \(\ln x \to -\infty\) as \(x \to 0^+\), the product \(x \ln x\) still approaches 0.
Important: The forms \(\infty - \infty\), \(1^\infty\), \(0^0\), and \(\infty^0\) are all indeterminate forms.
Strategy for \(\infty - \infty\): Combine terms into a single fraction to obtain \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) form.
Strategy for \(1^\infty\), \(0^0\), and \(\infty^0\): Take the natural logarithm of both sides, evaluate the limit of the logarithm (often converting to \(0 \cdot \infty\) form), then exponentiate the result.
Evaluate \(\displaystyle \lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1}\right)\)
Solution:
As \(x \to 0^+\), both \(\frac{1}{x} \to \infty\) and \(\frac{1}{e^x - 1} \to \infty\), giving the form \(\infty - \infty\). We combine into a single fraction:
\[\lim_{x \to 0^+} \left(\frac{1}{x} - \frac{1}{e^x - 1}\right) = \lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)}\]This has the form \(\frac{0}{0}\). Apply L'Hospital's Rule:
\[\lim_{x \to 0^+} \frac{e^x - 1 - x}{x(e^x - 1)} = \lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x}\]Still \(\frac{0}{0}\), apply L'Hospital's Rule again:
\[\lim_{x \to 0^+} \frac{e^x - 1}{e^x - 1 + xe^x} = \lim_{x \to 0^+} \frac{e^x}{e^x + e^x + xe^x} = \lim_{x \to 0^+} \frac{e^x}{e^x(2 + x)} = \frac{1}{1 + 1 + 0} = \frac{1}{2}\]Evaluate \(\displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^x\)
Solution:
As \(x \to \infty\), we have \(1 + \frac{3}{x} \to 1\) and the exponent \(x \to \infty\), giving the form \(1^\infty\). Let:
\[L = \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^x\]Take the natural logarithm of both sides:
\[\ln L = \lim_{x \to \infty} \ln\left[\left(1 + \frac{3}{x}\right)^x\right] = \lim_{x \to \infty} x \cdot \ln\left(1 + \frac{3}{x}\right)\]This has the form \(\infty \cdot 0\). Rewrite as a fraction:
\[\ln L = \lim_{x \to \infty} \frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{1}{x}}\]This has the form \(\frac{0}{0}\). Apply L'Hospital's Rule:
\[\ln L = \lim_{x \to \infty} \frac{\frac{1}{1 + \frac{3}{x}} \cdot \left(-\frac{3}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \to \infty} \frac{\frac{-3}{x^2(1 + \frac{3}{x})}}{-\frac{1}{x^2}}\]Simplifying:
\[\ln L = \lim_{x \to \infty} \frac{3}{1 + \frac{3}{x}} = \frac{3}{1 + 0} = 3\]Since \(\ln L = 3\), we have:
\[L = e^3\]Key Takeaway: When dealing with indeterminate forms of the type \(1^\infty\), \(0^0\), or \(\infty^0\):
L'Hospital's Rule Summary: