Question

Limits Depending on Direction of Approach $$\lim _{x \rightarrow 2^{-}} \frac{5+x}{x-2}$$

Short answer

\(\lim_{x \rightarrow 2^{-}} \frac{5+x}{x-2} = -\infty\)

Step-by-step solution

Understand the limit's direction

The notation 'x -> 2^-' indicates we are considering the limit as x approaches 2 from the left. This means we consider values of x that are slightly less than 2.

Analyze the function behavior as x approaches 2 from the left

Look at the denominator of the fraction (x - 2). As x gets very close to 2 from the left side, (x - 2) approaches zero and is negative. The numerator (5 + x) approaches 7 and is positive.

Determine the sign of the expression

Since the numerator approaches a positive value and the denominator approaches zero through negative values, the fraction's value becomes more negative, approaching negative infinity.

Conclude the limit

The limit of the function as x approaches 2 from the left side is negative infinity: \(\lim_{x \rightarrow 2^{-}} \frac{5+x}{x-2} = -\infty\).

Key concepts

Limit of a Function

Understanding the limit of a function is a fundamental concept in calculus. The limit describes the behavior of a function as the input (or the variable, typically represented by 'x') approaches a certain value. It's how we answer the question, 'What is the value of the function getting close to, as the input gets close to some number?' This does not necessarily mean the function will ever reach this value, but rather what value the function is approaching.

For example, in the limit \(\lim _{x \rightarrow 2^{-}} \frac{5+x}{x-2}\), we are interested in the value that the function \(\frac{5+x}{x-2}\) is approaching as 'x' gets very close to '2'. Importantly, the existence of a limit doesn't require the function to be defined at the point in question, it's all about the approach to that value.

Approaching from the Left

The approach from the left is denoted by the notation \(x \rightarrow a^{-}\). This tells us to look at values of 'x' that are less than 'a' but very close to it. Imagine moving along the graph towards 'a' from the left side.

In our exercise, we investigate \(x \rightarrow 2^{-}\), meaning we only consider 'x' values that are slightly less than '2'. Looking from the left allows us to understand how the function behaves as it gets close to the point from one specific direction. This is crucial because a function can behave differently from the left than it does from the right, leading to different one-sided limits.

Infinite Limits

An infinite limit occurs when the function's value increases or decreases without bound as the variable approaches a certain number. This often happens when a function has a vertical asymptote. An infinite limit is represented using the symbol \(\infty\) if the function grows without bound, or \( -\infty\) if the function decreases without bound.

In our case, as 'x' approaches '2' from the left, the denominator of our fraction \(x - 2\) is approaching zero, while the numerator is approaching a positive number, making the fraction grow negatively without bound. Therefore, the limit is \( -\infty\), which tells us that there is a vertical asymptote at 'x = 2', and as we approach from the left, the function descends precipitously.

Asymptotic Behavior

The term asymptotic behavior refers to how a function behaves as it moves closer to a line called an asymptote without actually reaching it. An asymptote is a straight line that a curve approaches as it heads towards infinity or minus infinity. Vertical asymptotes, in particular, occur at points where the function cannot be defined and where the limit results in an infinite value as the variable approaches a particular value.

Identifying Asymptotes

One way to identify these is by finding infinite limits. If as 'x' approaches a certain value the function grows or decreases without bounds as seen with infinite limits, a vertical asymptote is present.

In the exercise, the function \(\frac{5+x}{x-2}\) exhibits asymptotic behavior as 'x' approaches '2', because the function goes to negative infinity, implying a vertical asymptote at 'x = 2'.