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Chapter 2: Problem 42

Find the limits in Exercises \(37-48\) $$ \lim _{x \rightarrow-5^{-}} \frac{3 x}{2 x+10} $$

Short Answer

Expert verified
The limit is \(+\infty\) as \( x \to -5^{-} \).

Step by step solution

01

Understand the Problem

We need to find the limit of the function \( f(x) = \frac{3x}{2x + 10} \) as \( x \) approaches \(-5^{-}\). This means we are approaching -5 from the left side.
02

Substitute Directly and Simplify

Begin by substituting \( x = -5 \) directly into the expression to check for any indeterminate forms.\[ f(-5) = \frac{3(-5)}{2(-5) + 10} = \frac{-15}{-10 + 10} = \frac{-15}{0} \]The function gives us \( \frac{-15}{0} \), which is undefined, indicating a potential vertical asymptote at \( x = -5 \).
03

Analyze the Behavior as x Approaches from the Left

As \( x \) approaches \(-5\) from the left (\( x \to -5^{-} \)), the denominator \( 2x + 10 \) approaches 0 from the negative side since \( 2x + 10 = -10 + 10 \). This means as we approach \(-5\) from the left, \( 2x + 10 \) approaches 0 from the negative direction, resulting in a large negative value for the denominator.
04

Determine the Sign of the Limit

Since the numerator \( 3x \) equals \(-15\) as \( x \to -5 \), which is negative, and the denominator approaches a small negative number, the overall expression \( \frac{3x}{2x + 10} \) results in a positive value. As such, the limit approaches positive infinity as \( x \to -5^{-} \).
05

Conclusion of the Limit Calculation

Based on the preceding analyses, the limit is determined as follows:\[ \lim_{x \to -5^{-}} \frac{3x}{2x + 10} = +\infty \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

One-Sided Limits
One-sided limits are a crucial concept in calculus. They help us understand the behavior of functions as inputs approach a certain value from one side, either the left or right. In the expression \( \lim _{x \rightarrow-5^{-}} \frac{3 x}{2 x+10} \), the notation \(-5^{-}\) indicates that we are approaching -5 from the left. This is different from a two-sided limit, where we would approach the value from both sides.
  • One-sided limits are particularly useful when dealing with functions that have gaps or jumps at particular points.
  • They help us determine the behavior of a function near discontinuities.
Understanding one-sided limits enables us to make predictions about function values when direct substitution is not possible, such as when approaching a point of discontinuity like in our exercise.
Vertical Asymptotes
A vertical asymptote occurs at a point where a function's value becomes infinitely large or small as the input approaches some finite value. In the example \( \lim _{x \rightarrow-5^{-}} \frac{3 x}{2 x+10} \), substituting \( x = -5 \) makes the denominator zero, resulting in an undefined expression.
  • This is a tell-tale sign of a vertical asymptote.
  • Vertical asymptotes are represented visually as lines that a graph will approach but never touch.
As \( x \to -5^{-} \), the function \( \frac{3x}{2x+10} \) demonstrates a vertical asymptote because the denominator approaches zero, causing the fraction to diverge towards \( +\infty \). This asymptotic behavior is crucial for understanding complex graphs and functions.
Undefined Expressions
Undefined expressions occur when substituting a particular value results in an undefined mathematical form, like division by zero. In our exercise, substituting \( x = -5 \) into \( \frac{3x}{2x+10} \) leads to \( \frac{-15}{0} \), which is undefined.
  • Approaching such values often requires more advanced analysis, such as limits, to understand the function's behavior near these points.
  • Undefined expressions can indicate vertical asymptotes or removable discontinuities based on the context.
In calculus, when encountering undefined expressions, it's essential to explore the behavior of the function with limits to determine exact behavior as an input moves toward the undefined point.
Approaching from the Left
"Approaching from the left" is a term used to denote considering values of \( x \) that are slightly less than a given value. In the given expression \( \lim _{x \rightarrow-5^{-}} \frac{3 x}{2 x+10} \), \(-5^{-}\) indicates we are considering values of \( x \) that come from the left side of -5.
  • This situation often arises in functions with discontinuities or asymptotic behavior.
  • By understanding how the function behaves as it approaches from the left, we can determine the sign and value of the limit, even if the function is undefined at that point.
As \( x \) approaches \(-5\) from the left, the denominator approaches 0 from the negative side, drastically affecting the function's output, leading to a clear understanding that the output tends towards \( +\infty \) in this particular example.

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Most popular questions from this chapter

If \lim _{x \rightarrow 0} \frac{f(x)}{x^{2}}=1, find $$a. \lim _{x \rightarrow 0} f(x) \quad b. \lim _{x \rightarrow 0} \frac{f(x)}{x}$$

$$ \begin{array}{l}{\text { a. If } \lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=3, \text { find } \lim _{x \rightarrow 2} f(x)} \\ {\text { b. If } \lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=4, \text { find } \lim _{x \rightarrow 2} f(x)}\end{array} $$

Graph the curves in Exercises \(109-112 .\) Explain the relationship between the curve's formula and what you see. $$ y=\frac{x}{\sqrt{4-x^{2}}} $$

Find the limits \begin{equation} \lim \frac{x^{2}-3 x+2}{x^{3}-4 x} \end{equation} \begin{equation} \begin{array}{ll}{\text { a. }} & {x \rightarrow 2^{+}} \quad\quad\quad\quad {\text { b. } x \rightarrow-2^{+}} \\ {\text { c. }} & {x \rightarrow 0^{-}}\quad\quad\quad\quad {\text { d. } x \rightarrow 1^{+}} \\ {\text { e. }} & {\text { What, if anything, can be said about the limit as } x \rightarrow 0 ?}\end{array} \end{equation}

a. Graph \(h(x)=x^{2} \cos \left(1 / x^{3}\right)\) to estimate lim \(_{x \rightarrow 0} h(x),\) zooming in on the origin as necessary. b. Confirm your estimate in part (a) with a proof.
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