Question

$$\text { Prove that } \lim _{x \rightarrow \infty}\left(1+\frac{a}{x}\right)^{x}=e^{a}, \text { for } a \neq 0$$

Short answer

Question: Prove that when \(a\) is not equal to 0, the limit as\( x\) approaches infinity of \(\left(1+\frac{a}{x}\right)^{x}\) is equal to \(e^{a}\).

Step-by-step solution

Write down the limit

Begin by writing down the limit that we want to prove: $$\lim_{x \rightarrow \infty} \left(1+\frac{a}{x}\right)^{x}=e^{a}$$

Rewrite the limit using properties

Next, let's use the property of limits which states that if the limit of the function exists, then the limit of its natural logarithm exists as well. Define \(y(x) = \left(1+\frac{a}{x}\right)^{x}\). We can take the natural logarithm of both sides: $$\ln \left(\lim_{x \rightarrow \infty} \left(1+\frac{a}{x}\right)^{x}\right) = \ln \left(e^a\right)$$

Use definition of limit

Since the limit of the logarithmic function exists, apply the definition of limit so that we can perform the logarithm operation on the function: $$\lim_{x \rightarrow \infty} \ln\left(\left(1+\frac{a}{x}\right)^{x}\right) = \ln \left(e^a\right)$$

Simplify the limit

Use the properties of logarithms to simplify the limit expression. Since \({\ln{(e^a)}}=a\), this leaves us the following expression: $$\lim_{x \rightarrow \infty} \left(x \cdot \ln\left(1+\frac{a}{x}\right)\right) = a$$

Introduce a new variable

Let's introduce a new variable \(t\) and define it as \(t=\frac{a}{x}\). Consequently, \(x=\frac{a}{t}\). As \(x \rightarrow \infty\), then \(t \rightarrow 0\). Rewrite the limit using this substitution: $$\lim_{t \rightarrow 0} \left(\frac{a}{t} \cdot \ln\left(1+t\right)\right) = a$$

Use L'Hopital's Rule

Since the limit has the form \(\frac{0}{0}\), you can apply L'Hôpital's rule by taking derivatives of the numerator and denominator with respect to \(t\): $$\lim_{t \rightarrow 0} \frac{d}{dt}(a \cdot \ln(1+t))=\lim_{t \rightarrow 0} \frac{d}{dt}(t)$$ After taking derivatives, we get: $$\lim_{t \rightarrow 0} \frac{a \cdot \frac{1}{1+t}}{1}$$

Evaluate the limit

Now, evaluate the limit as \(t\) approaches 0: $$\frac{a \cdot \frac{1}{1+0}}{1}=a$$ We have proved that \(\lim_{t \rightarrow 0} a \cdot \ln\left(1+t\right) = a.\)

Reverse the natural logarithm step

Now that we have proven this limit, reverse the natural logarithm step from Step 3: $$\lim_{x \rightarrow \infty} \left(1+\frac{a}{x}\right)^{x}=e^{a}$$ This concludes the proof. We have shown that when \(a \neq 0\), the limit as \(x\) approaches infinity of \(\left(1+\frac{a}{x}\right)^{x}\) is equal to \(e^{a}\).

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule can be a great help when you're trying to solve limits that come in the indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). It provides a way to simplify these confusing expressions by using derivatives. Essentially, you take the derivative of the numerator and the derivative of the denominator, and then find the limit of their quotient.

In our problem, we faced an indeterminate form \( \frac{0}{0} \) when substituting directly. The expression involved the limit \( \lim_{t \rightarrow 0} \frac{a \cdot \ln(1+t)}{t} \) after the substitution \( t = \frac{a}{x} \).

Applying L'Hôpital's Rule required us to find the derivative of \( a \cdot \ln(1+t) \), which is \( \frac{a}{1+t} \), and the derivative of \( t \) which is simply 1. Evaluating the revised limit simplifies to \( \lim_{t \rightarrow 0} a \cdot \frac{1}{1+t} \).

Using this rule not only simplifies limits involving logarithms and rational expressions but also helps when these limits may appear daunting.

Natural Logarithm

The natural logarithm function, denoted as \( \ln(x) \), emerges frequently in calculus and is crucial for simplifying expressions involving exponents and limits. The natural logarithm has a base of \( e \), an irrational number approximately equal to 2.71828.

When dealing with expressions that involve powers, the natural logarithm can convert these into forms that are easier to manage using limit laws. In the given problem, we initially dealt with the exponential expression \( \left(1+\frac{a}{x}\right)^{x} \).

By taking the natural logarithm, we transformed it to a product of the form \( x \cdot \ln\left(1+\frac{a}{x}\right) \), which simplifies the handling of limits when \( x \to \infty \).

Knowing that \( \ln(e^a) = a \), the natural logarithm gives us access to directly compare expressions involving exponential limits by working through logarithms instead.

Limit Laws

Limit laws provide a foundation to evaluate the behavior of functions as they approach specific points or infinity. These laws offer tools to handle limits systematically and include rules like the sum, product, and quotient laws.

In our exercise, we used several limit laws to solve the expression \( \lim_{x \rightarrow \infty} \left(1+\frac{a}{x}\right)^{x} \). First, employing the property that if a limit exists for a function, it also exists for its natural logarithm allowed the transition from an exponential to logarithmic form.

Moreover, substituting \( t = \frac{a}{x} \) helped in exploring the limit through a simpler re-expressed variable as \( t \to 0 \). Simplifying this led us to find the derivative application needed for L’Hôpital’s Rule.

The combination of these laws helps establish continuity and ease in dealing with otherwise compounded expressions, thus confirming the limit result as \( e^{a} \).

Limit laws, when used skillfully, unravel complex looking expressions into manageable computations, facilitating precise problem solving.