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Chapter 2: Problem 12

Find the limits in Exercises \(11-18\) $$ \lim _{x \rightarrow 1^{+}} \sqrt{\frac{x-1}{x+2}} $$

Short Answer

Expert verified
The limit is 0.

Step by step solution

01

Identify the Type of Limit

The given limit is a one-sided limit as specified by the expression \( x \rightarrow 1^+ \), which means \( x \) is approaching 1 from the right. We'll examine how the function behaves as \( x \) approaches 1.
02

Substitute x = 1 in the function

Attempt to directly substitute 1 into the function: \( \sqrt{\frac{x-1}{x+2}} \). This gives \( \sqrt{\frac{1-1}{1+2}} = \sqrt{\frac{0}{3}} = \sqrt{0} = 0 \). This is straightforward, but we need to confirm that this substitution is valid by examining the behavior around \( x = 1 \).
03

Analyze Behavior of the Function Near x = 1

For values of \( x \) slightly greater than 1 (since we're considering \( 1^+ \)), \( x - 1 \) is a very small positive number, making \( \frac{x-1}{x+2} \) a very small positive quotient. Since the square root of a positive number is also positive, the function remains positive as \( x \) approaches 1 from the right.
04

Evaluate the Limit

Since there is no discontinuity or undefined behavior as \( x \rightarrow 1^+ \), the substitution made in Step 2 holds. Therefore, the limit is \( \lim_{x\rightarrow 1^+} \sqrt{\frac{x-1}{x+2}} = 0 \).

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Key Concepts

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

Limits
Limits in calculus are a fundamental concept used to investigate how a function behaves as the input approaches a particular value. They help us understand the behavior of functions, even at points where a direct substitution might not be possible. For example, limits can be employed when functions have undefined points or discontinuities. The notation \( \lim_{x \to a} f(x) \) denotes the value that \( f(x) \) approaches as \( x \) gets closer to \( a \).
  • **Finding Limits:** Often involves substituting the point into the function. However, when substitution gives an indeterminate form, other techniques like factoring or rationalizing might be used.
  • **Understanding Behavior:** Examining graphs or tables of values can also provide insight into what the function approaches.
To grasp limits, it's crucial to recognize their role in ensuring functions are evaluated appropriately, especially near points of interest that may not be straightforwardly defined.
One-Sided Limits
One-sided limits provide insight into the behavior of functions from specific directions - either approaching from the left (the negative side) or the right (the positive side). Notation like \( \lim_{x \to a^+} \) is used to describe the limit as \( x \) approaches \( a \) from the right.
  • **Right-Side Approach:** Denoted by \( x \to a^+ \), it considers values of \( x \) slightly greater than \( a \). This can reveal unique function characteristics hidden when only considering two-sided limits.
  • **Left-Side Approach:** Denoted by \( x \to a^- \), it concerns values slightly less than \( a \).
One-sided limits are particularly useful in analyzing situations where a function's behavior isn't symmetrical around \( a \). It allows for an in-depth look at gradual changes as the function transitions through a point.
Continuity
Continuity of a function at a point means the function is smooth and unbroken there. More formally, a function \( f(x) \) is continuous at \( x = a \) if
  • \( \lim_{x \to a} f(x) \) exists,
  • \( f(a) \) is defined,
  • And \( \lim_{x \to a} f(x) = f(a) \).
These criteria ensure there are no holes, jumps, or abrupt changes in the function at \( x = a \).
For a continuous function, one-sided limits from both directions will match and equal the function's value at \( a \). Discontinuities arise from violations of these conditions, such as limits not existing, differing one-sided limits, or the function not being defined at \( a \). Ensuring continuity often involves re-evaluating function definitions around problematic points.
Square Roots
The square root function, symbolized by \( \sqrt{x} \), provides a non-negative root of a number. When dealing with expressions within a function, such as \( \sqrt{\frac{x-1}{x+2}} \), understanding how square roots behave is crucial.
  • **Positivity:** The square root of a positive number remains positive, and \( \sqrt{0} \) is \( 0 \).
  • **Domain Considerations:** Ensure that the expression inside the square root is non-negative, as square roots of negative numbers are not real.
To work seamlessly with square roots in limit problems, it's important to properly assess the limit of expressions inside the root prior to evaluating the square root itself. Such precision ensures accurate limit calculations and avoids pitfalls linked with undefined or imaginary results.

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Most popular questions from this chapter

In Exercises \(19-22,\) find the slope of the curve at the point indicated. $$ y=1-x^{2}, \quad x=2 $$

In Exercises \(41-44,\) graph the function \(f\) to see whether it appears to have a continuous extension to the origin. If it does, use Trace and Zoom to find a good candidate for the extended function's value at \(x=0 .\) If the function does not appear to have a continuous extension, can it be extended to be continuous at the origin from the right or from the left? If so, what do you think the extended function's value(s) should be? $$ f(x)=(1+2 x)^{1 / x} $$

Does the graph of $$ g(x)=\left\\{\begin{array}{ll}{x \sin (1 / x),} & {x \neq 0} \\ {0,} & {x=0}\end{array}\right. $$ have a tangent at the origin? Give reasons for your answer.

Graphing Secant and Tangent Lines Use a CAS to perform the following steps for the functions in Exercises \(45-48 .\) a. Plot \(y=f(x)\) over the interval \(\left(x_{0}-1 / 2\right) \leq x \leq\left(x_{0}+3\right)\) b. Holding \(x_{0}\) fixed, the difference quotient $$ q(h)=\frac{f\left(x_{0}+h\right)-f\left(x_{0}\right)}{h} $$ at \(x_{0}\) becomes a function of the step size \(h .\) Enter this function into your CAS workspace. c. Find the limit of \(q\) as \(h \rightarrow 0\) d. Define the secant lines \(y=f\left(x_{0}\right)+q \cdot\left(x-x_{0}\right)\) for \(h=3,2\) and \(1 .\) Graph them together with \(f\) and the tangent line over the interval in part (a). $$ f(x)=x+\frac{5}{x}, \quad x_{0}=1 $$

Use a graphing calculator or computer grapher to solve the equations in Exercises \(63-70 .\) \(\cos x=x \quad\) (one root). Make sure you are using radian mode.
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