Problem 47 - Calculus for Biology and Medicine

Question

Find the limits in Problems 1-60; not all limits require use of l'Hôpital's rule. $$ \lim _{x \rightarrow \infty}\left(1-\frac{2}{x}\right)^{x} $$

Step-by-Step Solution

Answer

The limit is \( e^{-2} \).

Step 1: Recognize the Limit Form

The limit problem is \( \lim_{x \rightarrow \infty}\left(1-\frac{2}{x}\right)^{x} \). Observe that as \( x \to \infty \), \( 1 - \frac{2}{x} \to 1 \) and the expression is of the form \( 1^{\infty} \), an indeterminate form.

Step 2: Apply the Exponential Limit Formula

To solve an indeterminate form of \( 1^{\infty} \), use the formula \( \left(1 + \frac{a}{n}\right)^n \approx e^a \) as \( n \to \infty \). Rewrite the limit using this formula as it resembles the problem structure.

Step 3: Rewrite in Exponential Form

Rewrite the expression:\[\left(1 - \frac{2}{x}\right)^x = e^{x\ln\left(1 - \frac{2}{x}\right)}.\]

Step 4: Expand the Logarithm

Use the approximation \( \ln(1 - u) \approx -u \) for small \( u \). Thus,\[\ln\left(1 - \frac{2}{x}\right) \approx -\frac{2}{x}.\]

Step 5: Calculate the Limit of the Exponent

Substitute the approximation into the exponent:\[x \ln\left(1 - \frac{2}{x}\right) \approx x\left(-\frac{2}{x}\right) = -2.\]

Step 6: Evaluate the Exponential

Substitute back into the exponential form:\[e^{x \ln\left(1 - \frac{2}{x}\right)} \approx e^{-2}.\]Thus, the limit is \( e^{-2} \).

Key Concepts

l'Hôpital's rule

Understanding calculus limits often involves dealing with indeterminate forms. Suppose you have a limit like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These are indeterminate forms, and we need special tools to solve them. That's where l’Hôpital's Rule comes into play.
L'Hôpital's Rule is a method to find limits of indeterminate forms by differentiating the numerator and thedenominator. While our given problem doesn't need l'Hôpital's Rule, this is a powerful technique when faced with limits that appear stuck or difficult at first glance.
To use l'Hôpital's Rule, ensure:

• Your limit is in an indeterminate form of either \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \).
• Differentiate the numerator and the denominator separately.
• Calculate the limit of the new fraction.

By doing this, you can often simplify complex limit problems easily.

indeterminate forms

Indeterminate forms occur in calculus when the behavior of a limit is not immediately clear from its algebraic structure. In our original exercise, the expression \( \left(1 - \frac{2}{x}\right)^x \) approaches the form \( 1^\infty \) as \( x \to \infty \).
This "1 raised to infinity" is a classic indeterminate form. Such forms need special handling since they can lead to multiple potential limits rather than a clear single answer.
Common indeterminate forms include:

• \( \frac{0}{0} \)
• \( \frac{\infty}{\infty} \)
• \( 0 \times \infty \)
• \( \infty - \infty \)
• \( 1^\infty \), \( 0^0 \), and \( \infty^0 \)

Solving these forms often requires either algebraic manipulation, l'Hôpital's Rule, or techniques like the exponential limit formula, depending on the specific scenario.

exponential limit formula

When faced with an indeterminate form like \( 1^\infty \), one useful approach is the exponential limit formula.
For limits where the function resembles \( \left(1 + \frac{a}{n}\right)^n \) as \( n \) approaches infinity, the expression behaves like \( e^a \). This result is derived from the properties of the natural exponential function and can simplify the calculation of such limits.
In our exercise, we used this approach:

• Rewrote the expression in exponential form as \(e^{x\ln\left(1 - \frac{2}{x}\right)}\).
• Used the approximation \( \ln(1 - u) \approx -u \) for small \( u \) to transform it.
• Finally, calculated the limit of the exponent to simplify the expression to \( e^{-2} \).

This method can elegantly solve what initially seems a complicated limit problem by transforming it into something more familiar and manageable.