Problem 42 - A Graphical Approach to Precalculus with Limits

Question

Determine each limit, if it exists. $$\lim _{x \rightarrow 0}\left[2^{3 x}-\ln (x+1)\right]$$

Step-by-Step Solution

Answer

The limit is 1.

Step 1: Understand the Limit Expression

The limit we need to evaluate is \( \lim_{x \rightarrow 0} [2^{3x} - \ln(x + 1)] \). As \( x \) approaches 0, we need to understand the behavior of both terms in the expression: \( 2^{3x} \) and \( \ln(x + 1) \).

Step 2: Evaluate the Limit of the Exponential Term

Consider the term \( 2^{3x} \). As \( x \rightarrow 0 \), \( 3x \rightarrow 0 \) as well. We know that any number raised to the power of 0 equals 1, therefore, \( 2^{3x} \rightarrow 2^0 = 1 \).

Step 3: Evaluate the Limit of the Logarithmic Term

Next, evaluate \( \ln(x + 1) \). Notice that as \( x \rightarrow 0 \), \( x + 1 \) approaches 1. The natural logarithm of 1 is zero, so \( \ln(x + 1) \rightarrow \ln(1) = 0 \).

Step 4: Combine the Results

We combine the results from Steps 2 and 3. The limit of the entire expression \( 2^{3x} - \ln(x + 1) \) as \( x \rightarrow 0 \) becomes \( 1 - 0 = 1 \).

Step 5: Write the Final Answer

Thus, the limit of the expression as \( x \rightarrow 0 \) is \( 1 \).

Key Concepts

Exponential Functions

Exponential functions are mathematical functions of the form \(f(x) = a^x\), where \(a\) is a constant and \(x\) is an exponent. These functions have distinct characteristics:

• The base \(a\) is a positive real number not equal to \(1\).
• As the exponent \(x\) increases, the value of \(a^x\) rises rapidly for \(a > 1\), but decreases when \(0 < a < 1\).
• Exponential functions grow faster than polynomial and logarithmic functions for large values of \(x\).

In the context of limits, exponential functions can be easily evaluated. For example, \(2^{3x}\) in our exercise becomes \(2^0 = 1\) as \(x\) approaches 0, because

• Any non-zero number raised to the power 0 is 1, making the limit easy to compute.

Logarithmic Functions

Logarithmic functions are the inverse of exponential functions and have the form \(f(x) = \log_a (x)\), where \(a\) is the base. These functions have a unique set of properties that make them useful in various scenarios:

• They are the inverse of exponential functions, meaning if \(y = a^x\), then \(x = \log_a(y)\).
• Logarithms transform multiplication into addition, making complex calculations simpler.
• The natural logarithm \(\ln(x)\) is a special form of logarithm where the base \(a\) is \(e\), Euler's number, approximately equal to 2.71828.

For limits involving logarithms, it's crucial to understand their behavior near certain points. As seen with our exercise, \(\ln(x + 1)\) approaches \(\ln(1) = 0\) as \(x\) approaches 0. This is because the logarithm of 1 is always 0, which helps simplify limit calculations when the argument of \(\ln\) approaches values that are close to 1.

Limits and Continuity

Limits and continuity are foundational concepts in calculus, essential for understanding how functions behave as inputs approach certain values. A limit \(\lim_{x \to c} f(x)\) describes the value that \(f(x)\) approaches as \(x\) gets closer to \(c\). Here are some important things to know about limits:

• Limits can be one-sided, approaching from the left \((x \to c^-)\) or right \((x \to c^+)\).
• A limit exists only if the function approaches the same value from both sides.
• Indicator functions can also be used to evaluate limits, such as those involving infinity.

Continuity, on the other hand, means a function is unbroken or smooth. A function \(f(x)\) is continuous at a point \(c\) if:

• \(f(c)\) is defined.
• \(\lim_{x \to c} f(x)\) exists.
• \(\lim_{x \to c} f(x) = f(c)\).

In the given exercise, understanding limits allows us to determine that the expression \(2^{3x} - \ln(x + 1)\) approaches \(1\) as \(x\) approaches \(0\). Combining this knowledge with the properties of exponential and logarithmic functions aids in finding correct solutions.