/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 11 Find the limits in Exercises \(1... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the limits in Exercises \(11-18\) $$ \lim _{x \rightarrow-0.5^{-}} \sqrt{\frac{x+2}{x+1}} $$

Short Answer

Expert verified
The limit is \(\sqrt{3}\).

Step by step solution

01

Analyze the Expression

We start by examining the expression given in the problem: \( \sqrt{\frac{x+2}{x+1}} \). Our goal is to evaluate this function as \( x \) approaches \(-0.5\) from the left, which is indicated by the superscript \(^-\).
02

Simplify the Expression by Substitution

Substitute \( x = -0.5 \) into the functions inside the square root: \( \frac{x+2}{x+1} = \frac{-0.5 + 2}{-0.5 + 1} = \frac{1.5}{0.5} = 3 \). Note that \( x \) is approaching \(-0.5\) from the left, thus causing the fraction to approach 3.
03

Evaluate the Limit of the Root Function

Since the fraction approaches 3 as \( x \to -0.5^- \), the whole square root expression becomes \( \sqrt{3} \). Therefore, the limit of the original expression is \( \sqrt{3} \).

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Key Concepts

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

One-sided limit
A one-sided limit refers to the value a function approaches as the input approaches a specific point from one particular direction—either from the left or from the right. In mathematical terms, for the function in our problem, we are looking for the limit of \( \sqrt{\frac{x+2}{x+1}} \) as \( x \) approaches \(-0.5\) from the left.
  • The notation \( x \rightarrow -0.5^- \) indicates a left-handed limit.
  • In this problem, as \( x \) comes from values smaller than \(-0.5\), we only consider how the function behaves from that direction.
Taking a one-sided limit helps determine the behavior precisely at points where the function might have a different limit depending on the direction of approach, which is crucial in understanding limits comprehensively.It's especially important in cases of discontinuities or non-existent limits to clarify the nature of a limit condition. Thus, limits from one side, such as left or right, provide crucial insights for certain functions.
Square root function
The square root function is recognizable by the radical symbol \( \sqrt{ } \). It's a common function that appears in many types of mathematical expressions. For the function \( \sqrt{\frac{x+2}{x+1}} \), the square root operation modifies the value of the expression inside it.
  • It's important to remember that square roots are always non-negative when dealing with real numbers.
  • In this problem, \( \sqrt{3} \) is the end result after simplifying the fraction inside.
The squaring operation essentially transforms potentially negative results into non-negative ones, stressing the fact that the square root always produces results above or equal to zero in the context of real numbers.When we approximate such functions as part of limit calculations, it becomes crucial to accurately evaluate what's inside the square root first, as it directly influences the final outcome. This can often require additional algebraic manipulation.
Substitution method
The substitution method is a straightforward strategy used to simplify limit problems, especially when direct calculation is possible. This method involves substituting the value closest to the point of interest into the function to find a precise limit. Here's how it works in the given problem:
  • Considering the expression \( \frac{x+2}{x+1} \), we substitute \( x = -0.5 \).
  • With substitution, the expression becomes \( \frac{-0.5+2}{-0.5+1} = \frac{1.5}{0.5} = 3 \).
By substituting the exact value, you can examine the behavior of the function as the target value is approached closely, which leads to a more manageable expression. This technique is especially useful when dealing with algebraic fractions that might seem complex at first glance, simplifying them into more direct calculations. Moreover, this kind of substitution makes recognizing patterns or simplified expressions easier, paving the way towards calculating exact limits efficiently.

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Most popular questions from this chapter

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). $$ f(x)=x+\sin (2 x), \quad x_{0}=\pi / 2 $$

Find the limits in Exercises \(29-34 .\) Are the functions continuous at the point being approached? $$ \lim _{x \rightarrow \pi} \sin (x-\sin x) $$

In Exercises \(43-46\) , find a function that satisfies the given conditions and sketch its graph. (The answers here are are not unique. Any function that satisfies the conditions is acceptable. Feel free to use formulas defined in pieces if that will help.) $$ \lim _{x \rightarrow \pm \infty} f(x)=0, \lim _{x \rightarrow 2^{-}} f(x)=\infty, \text { and } \lim _{x \rightarrow 2^{+}} f(x)=\infty $$

Given \(\epsilon>0,\) find an interval \(I=(4-\delta, 4), \delta>0,\) such that if \(x\) lies in \(I,\) then \(\sqrt{4-x}<\epsilon .\) What limit is being verified and what is its value?

Circle's changing area What is the rate of change of the area of a circle \(\left(A=\pi r^{2}\right)\) with respect to the radius when the radius is \(r=3 ?\)

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