Question

Evaluate the following: \((a) \lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4}\) (b) \(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\) (c) \(\lim _{x \rightarrow 0} \frac{5^{x}-e^{x}}{x}\) \((d) \lim _{x \rightarrow \infty} \frac{\ln x}{x}\)

Short answer

(a) 7, (b) 1, (c) \(\ln 5 - 1\), (d) 0

Step-by-step solution

Problem (a) - Factor and Simplify

To evaluate the limit \( \lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4} \), first factor the numerator. The polynomial \( x^{2} - x - 12 \) can be factored as \((x-4)(x+3)\). The expression now becomes \(\frac{(x-4)(x+3)}{x-4}\). We can cancel out the \((x-4)\) term in the numerator and the denominator, leaving us with \(x+3\). Now evaluate the limit by substituting \(x = 4\): \(4 + 3 = 7\). Thus, the limit is 7.

Problem (b) - Apply L'Hôpital's Rule

For the limit \(\lim _{x \rightarrow 0} \frac{e^{x}-1}{x}\), recognize that the expression has the indeterminate form \(\frac{0}{0}\). Use L'Hôpital's Rule, which says we can differentiate the numerator and the denominator. The derivative of the numerator \(e^{x} - 1\) is \(e^{x}\), and the derivative of the denominator \(x\) is 1. The limit is now \(\lim _{x \rightarrow 0} e^{x}\), which equals \(e^{0} = 1\).

Problem (c) - Evaluate Using L'Hôpital's Rule

For \(\lim _{x \rightarrow 0} \frac{5^{x}-e^{x}}{x}\), check if it is in the indeterminate form \(\frac{0}{0}\). Both \(5^{x} - 1\) and \(e^{x} - 1\) approach 0 as \(x\) approaches 0, making the expression indeterminate. Apply L'Hôpital's Rule by differentiating the numerator and the denominator. The derivative of \(5^{x} - e^{x}\) is \(5^{x} \ln 5 - e^{x}\), and the derivative of \(x\) is 1. Thus, the new limit is \(\lim _{x \rightarrow 0} (5^{x} \ln 5 - e^{x})\). Substituting \(x = 0\), this limit simplifies to \(\ln 5 - 1\).

Problem (d) - Analyze Limit of logarithmic Function

To solve \(\lim _{x \rightarrow \infty} \frac{\ln x}{x}\), analyze the behavior of the function as \(x\) approaches infinity. Notice \(\ln x\) grows very slowly compared to \(x\), implying that the fraction \(\frac{\ln x}{x}\) heads towards zero. A more formal way to see this is to apply L'Hôpital's Rule: differentiate the numerator and denominator to get \(\lim _{x \rightarrow \infty} \frac{1/x}{1} = \lim _{x \rightarrow \infty} \frac{1}{x} = 0\), confirming that the limit is 0.

Key concepts

L'Hôpital's Rule

Imagine you face a tricky limit problem and end up with a form like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These are known as indeterminate forms, and that's exactly where L'Hôpital's Rule comes in handy. This rule helps us find limits for these specific situations by differentiating the numerator and the denominator.
For instance, consider the limit \( \lim _{x \rightarrow 0} \frac{e^{x}-1}{x} \). Here, directly substituting \(x = 0\) gives us \(\frac{0}{0}\), so it's indeterminate. According to L'Hôpital's Rule, we differentiate the top, yielding \(e^{x}\), and the bottom, giving us \(1\).
So the revised limit is \( \lim _{x \rightarrow 0} e^{x} = 1 \). This rule turns complex tasks simple, once understood.

Indeterminate Forms

Indeterminate forms are the puzzle pieces of limit evaluations. They occur when substituting a value into a limit expression results in an unclear answer, like \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These aren't immediately solvable because they can represent multiple values.
Let's look at the expression \( \lim _{x \rightarrow 0} \frac{5^{x} - e^{x}}{x} \). Plugging \(x = 0\) results in \(\frac{0}{0}\), an indeterminate form. Such cases require us to dig deeper to determine the actual limit.

• L'Hôpital's Rule becomes very useful for resolving indeterminate forms by allowing differentiation of the numerator and the denominator.
• By reducing the complexity, we can find meaningful results from seemingly impossible expressions.

Understanding and recognizing indeterminate forms are the keys to unlocking solutions to advanced calculus problems.

Polynomial Factorization

Polynomial factorization is like untangling a math knot. It involves breaking down a complex polynomial into simpler, multiplied components.
For example, consider the expression \(x^{2} - x - 12\), which needs factoring to simplify a limit problem: \( \lim _{x \rightarrow 4} \frac{x^{2}-x-12}{x-4} \).
Factor \(x^{2} - x - 12\) into \((x-4)(x+3)\) to reveal hidden simplicity. Canceling the \((x-4)\) terms from the numerator and denominator simplifies it to \(x + 3\).

• Immediate opportunity for substitution: Now, plug \(x = 4\) directly into \(x + 3\) to get the limit of 7.
• Factorization turns problems into easily solvable forms and is a useful tool in calculus.

This unveils a faster pathway to solutions, showing how factorization deconstructs challenges into approachable segments.

Logarithmic Functions

Logarithmic functions reveal growth trends and scale relationships. They grow slower than most polynomial functions.
In the exercise, the operation \( \lim _{x \rightarrow \infty} \frac{\ln x}{x} \) examines this trait head-on. Here, as \(x\) approaches infinity, \(\ln x\) increases, but not as swiftly as \(x\) itself.
This lets us suspect the limit approaches zero.

• Consider this intuitively: If one part grows much faster than the other, the quotient diminishes over time.
• To analyze more rigorously, apply L'Hôpital's Rule: Differentiate to obtain \( \lim _{x \rightarrow \infty} \frac{1/x}{1} = \lim _{x \rightarrow \infty} \frac{1}{x} = 0\).

Logarithmic functions often help model real-world phenomena, such as seismic scales or sound intensities, offering practical value in understanding growth dynamics.