Question

Find each limit. (a) \(\lim _{x \rightarrow 0^{+}}\left(1^{x}+2^{x}\right)^{1 / x}\) (b) \(\lim _{x \rightarrow 0^{-}}\left(1^{x}+2^{x}\right)^{1 / x}\) (c) \(\lim _{x \rightarrow \infty}\left(1^{x}+2^{x}\right)^{1 / x}\) (d) \(\lim _{x \rightarrow-\infty}\left(1^{x}+2^{x}\right)^{1 / x}\)

Short answer

(a) \(e\), (b) \(e\), (c) \(2\), (d) \(1\)

Step-by-step solution

Analyze expression for (a)

For \(x \to 0^+\), both \(1^x\) and \(2^x\) go to 1 respectively: \(1^x = 1\) and \(2^x \to 1^+\). We need to evaluate \(\lim_{x \to 0^+} (1^x + 2^x)^{1/x}\). Begin with the expression inside the parentheses: \((1^x + 2^x) = 1 + (1 + \epsilon)\). As \(x \to 0^+, \epsilon \to 0\).

Apply limit properties for (a)

Apply the limit to evaluate: \((1 + (1 + \epsilon))^{1/x}\). Since \(1 + \epsilon \to 1\), the expression turns into \((2 + \epsilon)^{1/x}\). Using: \((1 + \delta)^{1/\delta} \to e\) as \(\delta \to 0\), we compute \(\delta = x\ln(2+\epsilon)\), and \(\lim_{x \to 0^+} ((2+\epsilon)^{1/x}) = e\).

Analyze expression for (b)

For \(x \to 0^-\), \(1^x = 1\) and \(2^x \to 1^-\), the expression simplifies similarly: \(\lim_{x \to 0^-} ((1+\delta)^{1/x})\). The inner term becomes \((1 + (1 - \epsilon))\) where \(\epsilon \to 0\).

Apply limit properties for (b)

Taking the limit, the expression turns into \((2 - \epsilon)^{1/x}\). As \(\delta = x \ln(2 - \epsilon)\) approaches 0, similar logic to \(a\) shows \(\lim_{x \to 0^-} ((2 - \epsilon)^{1/x}) = e\).

Evaluate expression for (c)

For \(x \to \infty\), \(1^x = 1\) and \(2^x \to \infty\). The expression simplifies to \(( ext{large})^{1/x}\) meaning the base grows indefinitely, but any exponent \(1/x\) tends to zero. Thus the limit is: \(\lim_{x \to \infty} (2^x)^{1/x} = 2\).

Evaluate expression for (d)

For \(x \to -\infty\), \(1^x = 1\) and \(2^x \to 0\). Simplifying \((1 + \epsilon)^{1/x}\), where \(\epsilon \to 0^-\) leads to \(\lim_{x \to -\infty} 1^{1/x} = 1\).

Key concepts

Limit Evaluation

Evaluating limits is the process of determining the value that a function approaches as the input (variable) gets closer to a certain point. Understanding how to evaluate limits is a cornerstone of calculus, because it helps in analyzing the behavior of functions as they approach specific values, or infinity.

Here's a step-by-step guide to evaluating limits:

• **Identify the Approach**: Determine the values that the input variable is approaching, such as zero,infinity, or a positive or a negative sign.
• **Simplify Expressions**: Often, simplifying the function can make the process of evaluating thelimit easier.
• **Substitution of Limits**: Substitute the approaching values into the simpler expression to see howthe function behaves.

In cases where direct substitution leads to indeterminate forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\),other techniques such as L'Hopital's Rule or algebraic manipulation might be needed.

Understanding limit evaluation is important not only for solving calculus problems but also a crucial step in studying derivatives and integrals.

Exponential Functions

Exponential functions are mathematical expressions in which a constant base is raised to a variable exponent. These functions have the form \(a^x\), where \(a\) is the base and \(x\) is the exponent. They are widely used in calculus, always arousing interest due to their growth properties.

Key aspects of exponential functions include:

• **Growth and Decay**: Exponential functions can demonstrate both rapid increase (growth) or decrease (decay)depending on the value of the base \(a\).
• **Continuous and Smooth Graphs**: The graphs of exponential functions are continuous and smooth, making them distinctive and important for modeling real-world processes.
• **Logarithms Inverse**: The logarithm function is the inverse of an exponential function, providing tools for solving equations where the variable is an exponent.

Common examples of exponential functions include biological growth models, radioactive decay, and financial compounding.Understanding the behavior of these functions helps in evaluating complex limits as they demonstrate remarkable patterns when combined with limit operations like in the problems faced in our exercise.

Limit Properties

Limit properties are central characters in calculus that help simplify and solve complex limit problems. When dealing with limits, various properties make it easier to decompose expressions and compute values.

Some key limit properties include:

• **Sum Property**: The limit of a sum is equal to the sum of the individual limits, given both exist. Mathematically: \(\lim_{x \to a} (f(x) + g(x)) = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)\).
• **Product Property**: The limit of a product is the product of the limits. This helps in cases where limits of individual terms are easier to determine.
• **Exponential Growth Property**: For exponential functions, special properties such as \(\lim_{x \to \infty} (a^x)^{1/x} = a\) play a pivotal role in fast-tracking the evaluation of complicated expressions, as seen in the exercise where simplification occurred using this property.

Mastering these properties helps in systematic evaluation of limits, providing shortcuts to determine limitsthat would otherwise require complex calculations.