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Chapter 2: Problem 14

Find the limits. $$\lim _{x \rightarrow 1^{-}}\left(\frac{1}{x+1}\right)\left(\frac{x+6}{x}\right)\left(\frac{3-x}{7}\right)$$

Short Answer

Expert verified
The limit is 1.

Step by step solution

01

Understanding One-sided Limits

The problem asks us to find the limit of a function as \(x\) approaches \(1\) from the left, or \(x \to 1^-\). This means we need to consider values of \(x\) that are slightly less than \(1\).
02

Simplifying the Expression

First, simplify the entire expression: \( f(x) = \left(\frac{1}{x+1}\right)\left(\frac{x+6}{x}\right)\left(\frac{3-x}{7}\right)\). Multiplying these fractions gives: \(f(x) = \frac{(3-x)(x+6)}{7x(x+1)}\).
03

Substitute the Limit Approach Value

Since \(x \to 1^-\), substitute \(1\) into the simplified expression to check for any undefined values: \(f(1) = \frac{(3-1)(1+6)}{7(1)(1+1)} = \frac{(2)(7)}{14}\).
04

Compute the Expression

Continue from the substitution at Step 3: \(f(1) = \frac{14}{14} = 1\). Therefore, the expression evaluates to \(1\) when \(x\) is very close to \(1\) from the left side.

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Key Concepts

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

One-sided Limits
When working with limits in calculus, it's important to understand the concept of one-sided limits. A one-sided limit examines the behavior of a function as the input approaches a particular value from one specific direction, either from the left or the right. In this exercise, we are looking for the one-sided limit as \(x\) approaches \(1\) from the left, denoted as \(x \to 1^-\). This notation means we are considering values of \(x\) that are slightly less than \(1\).
  • To find a one-sided limit, you typically look at the values of \(x\) that are gradually getting closer to the target number from the left (or from the direction indicated).
  • If the function exhibits a different behavior approaching from the other side (\(x \to 1^+\)), you might get a different limit value.
  • Understanding the context and behavior of the function as \(x\) approaches the limit point is crucial to calculating one-sided limits accurately.
Limit Evaluation
The process of evaluating limits involves analyzing the function's behavior as it approaches a particular value. For one-sided limits, the focus is on just one side of the value in question.
To evaluate the limit of a function like the one in our example:
  • First, simplify the expression if possible to make it easier to evaluate.
  • Replace the approaching value into the simplified expression to see if it results in a real number or if it reveals an undefined behavior.
  • After substitution, compute the result to get the value of the limit as \(x\) approaches from the specified side. For our problem, substituting \(x = 1\) into the expression gives us the result \(1\), as seen in the solution.
Notice here that instead of getting an undefined form such as \(\frac{0}{0}\), our expression directly evaluated to a value without issue. Proper evaluation often hinges on this ability to simplify and substitute without error.
Fraction Simplification
Fraction simplification is a key step in many calculus problems, especially when evaluating limits. Simplifying fractions helps in making calculations easier and prevents errors during limit evaluation.
Key steps to simplify fractions in limits:
  • Multiply the numerators together and the denominators together to form a single fraction, as was done in the example: \(\frac{(3-x)(x+6)}{7x(x+1)}\).
  • Simplify complex expressions by canceling out common factors in the numerator and denominator if possible.
  • Make sure the simplification is valid for the limit point \(x\) approaches. Sometimes adjustments are needed if the point causes division by zero or undefined terms.
In this problem, simplifying before substituting the approaching value made it easier to verify whether the function would result in an indeterminate form. Successful simplification allows for clear substitution and result computation.

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Most popular questions from this chapter

Gives a function \(f(x)\) and numbers \(L, c,\) and \(\epsilon \geq 0 .\) In each case, find an open interval about \(c\) on which the inequality \(|f(x)-L|<\epsilon\) holds. Then give a value for \(\delta>0\) such that for all \(x\) satisfying \(0<|x-c|<\delta\) the inequality \(|f(x)-L|<\epsilon\) holds. $$f(x)=m x, \quad m>0, \quad L=2 m, \quad c=2, \quad \epsilon=0.03$$

a. If \(\lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=3,\) find \(\lim _{x \rightarrow 2} f(x).\) b. If \(\lim _{x \rightarrow 2} \frac{f(x)-5}{x-2}=4,\) find \(\lim _{x \rightarrow 2} f(x).\)

A fixed point theorem Suppose that a function \(f\) is continuous on the closed interval [0,1] and that \(0 \leq f(x) \leq 1\) for every \(x\) in \([0,1] .\) Show that there must exist a number \(c\) in [0,1] such that \(f(c)=c(c \text { is called a fixed point of } f)\)

Use a CAS to perform the following steps: a. Plot the function \(y=f(x)\) near the point \(c\) being approached. b. Guess the value of the limit \(L\) and then evaluate the limit symbolically to see if you guessed correctly. c. Using the value \(\epsilon=0.2,\) graph the banding lines \(y_{1}=L-\epsilon\) and \(y_{2}=L+\epsilon\) together with the function \(f\) near \(c\) d. From your graph in part (c), estimate a \(\delta>0\) such that for all \(x\) \(0<|x-c|<\delta \Rightarrow|f(x)-L|<\epsilon\) Test your estimate by plotting \(f, y_{1},\) and \(y_{2}\) over the interval \(0<|x-c|<\delta .\) For your viewing window use \(c-2 \delta \leq\) \(x \leq c+2 \delta\) and \(L-2 \epsilon \leq y \leq L+2 \epsilon .\) If any function values lie outside the interval \([L-\epsilon, L+\epsilon],\) your choice of \(\delta\) was too large. Try again with a smaller estimate. e. Repeat parts (c) and (d) successively for \(\epsilon=0.1,0.05,\) and 0.001 $$f(x)=\frac{3 x^{2}-(7 x+1) \sqrt{x}+5}{x-1}, \quad c=1$$

You will find a graphing calculator useful for Exercise. Let \(f(x)=\left(x^{2}-9\right) /(x+3).\) a. Make a table of the values of \(f\) at the points \(x=-3.1\) \(-3.01,-3.001,\) and so on as far as your calculator can go. Then estimate \(\lim _{x \rightarrow-3} f(x) .\) What estimate do you arrive at if you evaluate \(f\) at \(x=-2.9,-2.99,-2.999, \ldots\) instead? b. Support your conclusions in part (a) by graphing \(f\) near \(c=-3\) and using Zoom and Trace to estimate \(y\) -values on the graph as \(x \rightarrow-3.\) c. Find \(\lim _{x \rightarrow-3} f(x)\) algebraically, as in Example 7.
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