Question

Limits Evaluate the following limits. Use l'Hópital's Rule when it is comvenient and applicable. $$\lim _{z \rightarrow \infty}\left(1+\frac{10}{z^{2}}\right)^{z^{2}}$$

Short answer

Answer: The limit of the expression is \(e^{10}\).

Step-by-step solution

Rewrite the expression using natural logarithm

First, we will apply natural logarithm (ln) to the expression so that we could bring down the exponent and use exponential limit formula. For that, let L be our limit: $$L = \lim_{z \rightarrow \infty}\left(1+\frac{10}{z^{2}}\right)^{z^{2}}$$ Now, apply the natural logarithm: $$\ln{L} = \ln{\left[\lim_{z\rightarrow\infty}\left(1+\frac{10}{z^{2}}\right)^{z^{2}}\right]} = \lim_{z\rightarrow\infty} z^{2} \cdot \ln{\left(1+\frac{10}{z^{2}}\right)}$$

Apply L'Hôpital's Rule

Now, we have an indeterminate form and can use L'Hôpital's Rule. However, the limit is in the form of a product, not a fraction. We will rewrite the expression in the form of a fraction: $$\ln{L} = \lim_{z\rightarrow\infty} \frac{\ln{\left(1+\frac{10}{z^{2}}\right)}}{\frac{1}{z^2}}$$ In this form, both the numerator and the denominator approach zero as z approaches infinity, so we can apply l'Hôpital's Rule. Differentiating the numerator with respect to z: $$\frac{d}{dz}\left[\ln{\left(1+\frac{10}{z^2}\right)}\right] = \frac{-20z^{-3}}{1 + \frac{10}{z^2}}$$ Differentiating the denominator with respect to z: $$\frac{d}{dz}\left[\frac{1}{z^2}\right] = -2z^{-3}$$ So, after applying l'Hôpital's Rule, we have: $$\ln{L} = \lim_{z\rightarrow\infty} \frac{\frac{-20z^{-3}}{1 + \frac{10}{z^2}}}{-2z^{-3}}$$

Simplify the expression after applying L'Hôpital's Rule

Simplifying the expression and canceling out the common terms: $$\ln{L} = \lim_{z\rightarrow\infty} \frac{-20}{-2\left(1 + \frac{10}{z^2}\right)} = \frac{10}{1+0} = 10$$ Now, we've found the value of \(\ln{L}\), and we need to find the value of L. Take the exponential of both sides, $$L = e^{10}$$ So, the final answer of the limit is: \(\lim_{z \rightarrow \infty}\left(1+\frac{10}{z^{2}}\right)^{z^{2}} = e^{10}\)

Key concepts

Limits

In calculus, limits help us understand the behavior of functions as they approach specific points, typically as the input values approach infinity or a particular number. In our example, we are interested in understanding what happens to a function of the form \( \left(1 + \frac{10}{z^2}\right)^{z^2} \) as \( z \) approaches infinity. Limits provide a way to explore concepts such as rates of change and continuity.When evaluating limits, it often involves substituting values and simplifying expressions to observe their behavior. In cases where direct substitution leads to indeterminate forms, further techniques like L'Hôpital's Rule become necessary. Understanding how to navigate limits forms the foundation of more advanced calculus techniques.

L'Hôpital's Rule

L'Hôpital's Rule is a powerful tool in calculus used when evaluating limits that result in indeterminate forms, such as \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It simplifies the process by allowing us to differentiate the numerator and the denominator separately, making the limit easier to evaluate.

Applying L'Hôpital's Rule

• First, identify the indeterminate form in the limit expression.
• Differentiate the numerator and denominator separately.
• Evaluate the limit of the resulting expression.

In our problem, after taking the natural logarithm, we recognize an indeterminate form and reform the limit as a fraction: \( \frac{\ln(1 + \frac{10}{z^2})}{1/z^2} \). By applying L'Hôpital's Rule, we differentiate the numerator and denominator. This simplification reveals a limit that's straightforward to solve versus the original expression.

Indeterminate Forms

Indeterminate forms arise when substituting into a limit results in expressions like \( \frac{0}{0} \), \( \frac{\infty}{\infty} \), \( 0\cdot\infty \), among others. These forms do not provide enough information to determine the limit outright. Therefore, additional algebraic or analytic manipulation is required to find a meaningful result.

Examples and Solutions

• \( \frac{0}{0} \): Common in rational functions and solvable by factorization or L'Hôpital's Rule.
• \( \infty - \infty \): Requires transformation or adjusting terms to find a limit.
• \( 1^\infty \): Seen in expressions involving exponents, like our exercise. Often handled with logarithms or the exponential limit formula.

Indeterminate forms necessitate techniques like L'Hôpital's Rule, which leverages calculus principles to transform unsolvable limits into evaluable expressions. This process broadens the tools available for finding limits, accommodating a wider range of functions and scenarios in calculus.