Question

Evaluate the following limits in terms of the parameters a and b, which are positive real numbers. In each case, graph the function for specific values of the parameters to check your results. $$\lim _{x \rightarrow 0}(1+a x)^{b / x}$$

Short answer

$$\lim_{x \rightarrow 0} (1+ax)^{b/x}$$ Answer: The limit of the given expression is: $$e^{ab}$$

Step-by-step solution

Rewrite the expression using natural logarithm and exponential functions

We will rewrite the expression using natural logarithm and exponential functions as follows: $$\lim_{x \rightarrow 0} (1+ax)^{b/x}=\lim_{x \rightarrow 0} e^{\ln((1+ax)^{b/x})}$$

Simplify the exponent

Next, we will simplify the exponent using logarithm properties: $$\lim_{x \rightarrow 0} e^{\ln((1+ax)^{b/x})} = \lim_{x \rightarrow 0} e^{(b/x)\cdot \ln(1+ax)}$$

Define a new variable and rewrite the limit

Let \(y = (b/x)\cdot \ln(1+ax)\). Our limit will now be of the form: $$\lim_{y \rightarrow 0} e^{y}$$ To find the limit for this new expression, we will first find the limit for y as x approaches 0.

Write the expression for y

The expression for y is: $$y = \frac{b\ln(1+ax)}{x}$$

Apply L'Hopital's Rule

The expression for y has an indeterminate form \((0/0)\). We'll apply L'Hopital's Rule to find the limit of this expression as x approaches 0: $$\lim_{x \rightarrow 0} \frac{b\ln(1+ax)}{x} = \lim_{x \rightarrow 0} \frac{b\frac{a}{1+ax}}{1}$$ Now, as x approaches 0, the limit becomes: $$\lim_{x \rightarrow 0} b\cdot\frac{a}{1+a\cdot0} = b\cdot\frac{a}{1} = ab$$

Find the limit of the original expression

Now that we have the limit of y as x approaches 0, we can find the limit of the original expression: $$\lim_{y \rightarrow 0} e^{y} = e^{ab}$$ The limit of the given expression, with parameters a and b as positive real numbers, is: $$\lim_{x \rightarrow 0}(1+ax)^{b/x} = e^{ab}$$

Key concepts

L'Hopital's Rule

L'Hopital's Rule is an essential tool in calculus to evaluate limits that result in indeterminate forms. These forms often appear in fractions that evaluate to \(0/0\) or \(\infty/\infty\). When you encounter such cases, you can apply L'Hopital's Rule to find the limit more easily.
To use L'Hopital's Rule, follow these steps:

• First, confirm that your limit indeed results in an indeterminate form like \(0/0\) or \(\infty/\infty\).
• Next, differentiate the numerator and the denominator separately.
• Then, calculate the limit of the new fraction.

It's important that the new limit should not be indeterminate. If it is, you might have to repeat the process. In the problem we initially have \(\lim_{x \rightarrow 0} \frac{b\ln(1+ax)}{x}\), which is in \(0/0\) form. By differentiating, we find the limit as \(ab\), making the original expression solvable.
Remember, L'Hopital's Rule only applies to certain indeterminate forms, so it’s crucial to verify this first before applying it.

Natural Logarithm

Natural logarithms are logarithms to the base \(e\), which is an irrational number approximately equal to 2.71828. They are a key concept in calculus and are used extensively to transform expressions in beneficial ways.
Natural logarithms have several important properties:

• \(\ln(1) = 0\): The logarithm of 1 is always zero.
• \(\ln(e) = 1\): The logarithm of \(e\) is one.
• Logarithms turn multiplication into addition, as in \(\ln(xy) = \ln(x) + \ln(y)\).
• Similarly, \(\ln(x^y) = y\ln(x)\), which helps simplify powers, as seen in the problem where \(\ln((1+ax)^{b/x})\) was simplified to \((b/x)\cdot \ln(1+ax)\).

In the exercise, the natural logarithm is used to rewrite the function into a form that makes applying L'Hopital's Rule possible. It allows complex expressions involving exponents to be transformed and evaluated more directly.

Exponential Functions

Exponential functions are functions of the form \(f(x) = a \cdot e^{bx}\), where \(e\) is Euler's number. They model many natural processes and have distinct properties that make them ideal for calculus.
Key characteristics of exponential functions include:

• The function \(e^x\) grows rapidly as \(x\) increases and never touches the x-axis, staying above it.
• Exponential functions are their own derivatives and integrals, making them particularly valuable in calculus.
• They are used to "undo" logarithmic functions, helping solve equations involving logarithms.

In our problem, after converting the original expression using natural logarithms, we find the limit using exponential expressions as \(e^y\). The simplicity of exponential functions allows for straightforward computation of limits, such as \(e^{ab}\), once the exponent \(y\) is determined.