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Chapter 2: Problem 78

Evaluate the following limits. \(\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{4 x+5}-3}\)

Short Answer

Expert verified
Question: Evaluate the limit as x approaches 1 for the expression $\frac{x-1}{\sqrt{4x+5}-3}$. Answer: The limit as x approaches 1 for the given expression is $\frac{3}{2}$.

Step by step solution

01

Identify the indeterminate form

Plug in the value of \(x=1\) in the given expression: $$\frac{1-1}{\sqrt{4(1)+5}-3} = \frac{0}{0}$$ Since we get an indeterminate form, we need to simplify the expression before we can evaluate the limit.
02

Rationalize the denominator

Multiply the numerator and denominator of the expression by the conjugate of the denominator: $$\frac{x-1}{\sqrt{4 x+5}-3} \times \frac{\sqrt{4 x+5}+3}{\sqrt{4 x+5}+3} = \frac{(x-1)(\sqrt{4 x+5}+3)}{((4 x+5)-9)}$$
03

Simplify the expression

Simplify the numerator and denominator of the expression: $$\frac{(x-1)(\sqrt{4 x+5}+3)}{(4 x-4)} = \frac{(x-1)(\sqrt{4 x+5}+3)}{4(x-1)}$$ Now, cancel out \((x-1)\) from the numerator and the denominator: $$\frac{\sqrt{4 x+5}+3}{4}$$
04

Evaluate the limit

Now that the expression is simplified, plug in the value \(x=1\) to evaluate the limit: $$\lim_{x \to 1} \frac{\sqrt{4 x+5}+3}{4} = \frac{\sqrt{4(1)+5}+3}{4} = \frac{\sqrt{9}+3}{4} = \frac{6}{4} = \frac{3}{2}$$ Therefore, the limit is: $$\lim _{x \rightarrow 1} \frac{x-1}{\sqrt{4 x+5}-3} = \frac{3}{2}$$

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Key Concepts

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

Rationalize the Denominator
Rationalizing the denominator is an essential algebraic technique used to eliminate irrational numbers from under the square root in the denominator. This process makes evaluating limits and simplifying expressions much easier. In the given problem, we encounter a square root in the denominator, which prompts us to rationalize it.

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of a binomial expression \(a - b\) is \(a + b\). Thus, in this problem, the conjugate of \(\sqrt{4x+5} - 3\) is \(\sqrt{4x+5} + 3\).

When you multiply by the conjugate, the denominator becomes \( ((4x+5) - 9) \), which simplifies it due to the difference of squares, eliminating the square root. This rationalization transforms the given expression into a form where the indeterminate \(\frac{0}{0}\) can be resolved more easily.
Indeterminate Form
Indeterminate forms are expressions that do not initially possess a defined value but can be reworked to find a meaningful result. They often occur in calculating limits, where a straightforward substitution results in forms such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). These forms signal the need for algebraic manipulation to resolve the expression.

In our problem, substituting \(x = 1\) into the function yields \(\frac{0}{0}\), an indeterminate form. This outcome means the limit can not be directly solved and requires further operations to simplify the expression into a determinate form. Identifying these scenarios early lets us know that simplification steps, such as rationalization, are necessary, ensuring we correctly evaluate the limit.
Simplifying Expressions
Simplifying expressions is a critical step in calculating limits, especially when dealing with complicated mathematical expressions. After rationalizing the denominator, our expression still needs to be simplified before the limit can be accurately evaluated.

In this exercise, after multiplying by the conjugate, we're left with a complex fraction. We notice that the expression \((x-1)\) appears both in the numerator and the denominator. This common factor can be canceled out, leaving a simplified expression of the form \(\frac{\sqrt{4x+5} + 3}{4}\). This new form is no longer in indeterminate form, allowing for substitution to solve the limit.

By simplifying expressions, complex problems become more manageable, revealing their essential components, and enabling accurate solutions like determining the limit in this exercise.

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Most popular questions from this chapter

Evaluate the following limits or state that they do not exist. $$\lim _{x \rightarrow 0^{+}} \frac{x}{\ln x}$$

Use the following definition for the nonexistence of a limit. Assume \(f\) is defined for all values of \(x\) near a, except possibly at a. We write \(\lim _{x \rightarrow a} f(x) \neq L\) if for some \(\varepsilon>0\) there is no value of \(\delta>0\) satisfying the condition $$|f(x)-L|<\varepsilon \quad \text { whenever } \quad 0<|x-a|<\delta.$$ Let $$f(x)=\left\\{\begin{array}{ll} 0 & \text { if } x \text { is rational } \\ 1 & \text { if } x \text { is irrational. } \end{array}\right.$$ Prove that \(\lim _{x \rightarrow a} f(x)\) does not exist for any value of \(a\). (Hint: Assume \(\lim _{x \rightarrow a} f(x)=L\) for some values of \(a\) and \(L\) and let \(\varepsilon=\frac{1}{2}\).)

Evaluate the following limits, where \(c\) and \(k\) are constants. \(\lim _{h \rightarrow 0} \frac{100}{(10 h-1)^{11}+2}\)

Calculate the following limits using the factorization formula \(x^{n}-a^{n}=(x-a)\left(x^{n-1}+x^{n-2} a+x^{n-3} a^{2}+\cdots+x a^{n-2}+a^{n-1}\right)\) where \(n\) is a positive integer and a is a real number. $$\lim _{x \rightarrow 1} \frac{\sqrt[3]{x}-1}{x-1}\left(\text { Hint: } x-1=(\sqrt[3]{x})^{3}-(1)^{3}\text { ). }\right.$$

Find the limit of the following sequences or state that the limit does not exist. $$\begin{aligned} &\left\\{\frac{1}{2}, \frac{4}{3}, \frac{9}{4}, \frac{16}{5}, \ldots\right\\}, \text { which is defined by } f(n)=\frac{n^{2}}{n+1}, \text { for }\\\ &n=1,2,3, \dots \end{aligned}$$
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