Question

Prove the following limit relations: (a) \(\lim _{x \rightarrow 0} \frac{b^{x}-1}{x}=\log b \quad(b>0)\). (b) \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\). (c) \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\). (d) \(\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}\).

Short answer

Question: Find the limits of the following expressions: a) \(\lim _{x \rightarrow 0} \frac{b^{x}-1}{x}\) b) \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}\) c) \(\lim _{x \rightarrow 0}(1+x)^{1 / x}\) d) \(\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}\) Answer: a) \(\lim _{x \rightarrow 0} \frac{b^{x}-1}{x} = \log b\) b) \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x} = 1\) c) \(\lim_{x \rightarrow 0}(1+x)^{1/x} = e\) d) \(\lim_{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n} = e^x\)

Step-by-step solution

Apply logarithmic property

First, recall the property \((\log_a b)^c = \log_a(b^c)\). Rewrite the limit as: \[\lim _{x \rightarrow 0} \frac{\log_b (b^x)-\log_b(1)}{x}.\]

Use L'Hôpital's rule

Now apply L'Hôpital's rule to find the limit: \[\lim _{x \rightarrow 0} \frac{\log_b (b^x)'}{1} = \left(\log_b (\mathrm{d}(b^x) / \mathrm{d}x)\right)_{x=0} = \left(\log_b (b^x \ln b)\right)_{x=0} = \log_b(\ln b) = \log b.\] (b) Proving \(\lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\)

Change of variables

Let \(y = 1+x\). Then, \(x = y-1\), and as \(x \rightarrow 0\), \(y \rightarrow 1\). The limit becomes: \[\lim _{y \rightarrow 1} \frac{\log y}{y-1}.\]

Use L'Hôpital's rule

Applying L'Hôpital's rule: \[\lim _{y \rightarrow 1} \frac{\log y'}{(y-1)'} = \lim_{y \rightarrow 1} \frac{1/y}{1} = 1.\] (c) Proving \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e\)

Use natural logarithm

Take the natural logarithm of the expression: \[\ln\left((1+x)^{1 / x}\right) = \frac{1}{x} \ln(1+x).\]

Use L'Hôpital's rule

Now, we will find the limit of this expression as \(x \rightarrow 0\) using L'Hôpital's rule: \[\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x} = \lim_{x \rightarrow 0} \frac{\log(1+x)'}{1} = \lim_{x \rightarrow 0} \frac{1}{1+x} = 1.\]

Exponentiate

Since we took the natural logarithm at the beginning, we must exponentiate to find the desired limit: \[\lim_{x \rightarrow 0}(1+x)^{1/x} = e^{\lim_{x \rightarrow 0} \frac{\ln(1+x)}{x}} = e^1 = e.\] (d) Proving \(\lim _{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n}=e^{x}\)

Use natural logarithm

Take the natural logarithm of the expression: \[\ln\left(\left(1+\frac{x}{n}\right)^{n}\right) = n\ln\left(1+\frac{x}{n}\right).\]

Use L'Hôpital's rule

Now, we will find the limit of this expression as \(n \rightarrow \infty\) using L'Hôpital's rule: \[\lim_{n \rightarrow \infty} \frac{n \ln(1+x/n)}{x} = \lim_{n \rightarrow \infty} \frac{\ln(1+x/n)}{x/n} = \lim_{n \rightarrow \infty} \frac{\log(1+x/n)'}{(-x)/n^2} = \lim_{n \rightarrow \infty} \frac{-n^2}{x(1+x/n)} = x.\]

Exponentiate

Since we took the natural logarithm at the beginning, we must exponentiate to find the desired limit: \[\lim_{n \rightarrow \infty}\left(1+\frac{x}{n}\right)^{n} = e^{\lim_{n \rightarrow \infty} n\ln(1+x/n)} = e^x.\]

Key concepts

L'Hôpital's Rule

L'Hôpital's Rule is a fundamental concept in calculus used to find the limit of indeterminate forms, such as \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). This rule states that if the limit of a function results in an indeterminate form, you can differentiate the numerator and the denominator independently, and then take the limit of their quotient.
Here’s how it works:

• First, check if the initial limit expression is in an indeterminate form.
• Differentiate the function in the numerator and the function in the denominator separately.
• Calculate the limit of the new fraction.
• If it is still indeterminate, L'Hôpital's Rule can be applied repeatedly until a determinable limit is found.

This rule is particularly useful when dealing with limits involving exponential and logarithmic functions, as it can simplify calculations and provide clear solutions when direct substitution or algebraic manipulation fails. See how it is applied to exponential and logarithmic functions in the solution steps.

Exponential Functions

Exponential functions are mathematical expressions in the form \( a^x \), where \( a \) is a positive constant and \( x \) can be any real number. These functions are characterized by constant rates of change and are widely used to model situations of growth or decay.
Some key properties of exponential functions include:

• The base \( a \) must always be a positive number, and cannot be 1, as this makes the function constant.
• The graph of an exponential function is a curve that increases (or decreases) rapidly, reflecting growth or decay.
• They possess the key property that the derivative of \( a^x \) is proportional to \( a^x \) itself.

Exponential functions often appear in limits and are sometimes best handled using logarithms or L'Hôpital's Rule to determine limit values, as seen in the original step by step solution. Understanding their behavior is essential in solving many mathematical and practical problems.

Logarithmic Functions

A logarithm is the inverse process of an exponential function. Given an exponential function \( a^x = y \), a logarithm finds \( x \) as \( \log_a y \). Here are some critical aspects of logarithmic functions:

• The base logarithm \( \log_a \) asks "To what power should the base \( a \) be raised, to produce \( y \)?"
• Common logs include the natural log, \( \ln \), which uses base \( e \), and the base 10 log, \( \log_{10} \).
• Logarithms transform multiplicative relationships into additive ones, making them a valuable tool for simplifying complex equations.
• The derivative of \( \ln(x) \) is \( \frac{1}{x} \), making it easier to calculate the limits of logarithmic expressions.

Logarithmic functions are commonly paired with exponential expressions, especially in limit problems. By converting exponentials to logarithms, you can often simplify the problem enough to apply L'Hôpital's Rule and find the limit more effectively. This is efficiently illustrated in part of the original solution process. Understanding logarithmic behavior is crucial in mastering advanced mathematical concepts.