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Find the limit (if it exists). \(\lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x}\)

Short Answer

Expert verified
The limit of the given function as \(\Delta x\) approaches 0 is \(\frac{1}{2\sqrt{x+2}}\).

Step by step solution

01

Multiply by the conjugate

To begin, multiply the numerator and denominator by the conjugate of the numerator: \(\sqrt{x+2+\Delta x} + \sqrt{x+2}\). This gives us: \[\lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x} \cdot \frac{\sqrt{x+2+\Delta x} + \sqrt{x+2}}{\sqrt{x+2+\Delta x} + \sqrt{x+2}}\]
02

Simplify the multiplication

Simplify the multiplication. This should give us a difference of squares in the numerator, that can be simplified into: \[ \lim _{\Delta x \rightarrow 0} \frac{(\Delta x)}{\Delta x (\sqrt{x+2+\Delta x} + \sqrt{x+2})}\]
03

Cancel the term \(\Delta x\)

The term \(\Delta x\) in the numerator and denominator can be cancelled, which should simplify our expression into: \[ \lim _{\Delta x \rightarrow 0} \frac{1}{\sqrt{x+2+\Delta x} + \sqrt{x+2}}\]
04

Apply the limit

Now that the indeterminate form is removed, apply the limit by substituting \(\Delta x = 0\), which gives: \[\frac{1}{\sqrt{x+2+0} + \sqrt{x+2}} = \frac{1}{2\sqrt{x+2}}\]

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Key Concepts

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

Difference of Squares
In calculus, understanding the difference of squares is important, especially when evaluating limits involving square roots. The difference of squares formula is expressed as \( a^2 - b^2 = (a - b)(a + b) \). This formula is useful because it allows us to simplify expressions that involve subtraction between two squared terms. When dealing with expressions like \( \sqrt{x+2+\Delta x} - \sqrt{x+2} \), multiplying by the conjugate helps in forming a difference of squares. The conjugate in this scenario is \( \sqrt{x+2+\Delta x} + \sqrt{x+2} \). Using the difference of squares formula simplifies this expression to \( (a-b)(a+b) = a^2 - b^2 \), which is easier to manage during limit evaluation.
Conjugate Method
The conjugate method is a powerful technique in calculus for simplifying the expressions involving square roots. It helps eliminate radicals or roots in complex expressions. In this context, a conjugate is essentially the same two terms with opposite signs. For example:
  • The expression \( \sqrt{x+2+\Delta x} - \sqrt{x+2} \) has a conjugate: \( \sqrt{x+2+\Delta x} + \sqrt{x+2} \).
When you multiply by the conjugate, it assists in eliminating the radicals by forming a difference of squares. This action results in terms that are easier to handle algebraically, which is essential in solving limit problems and tackling indeterminate forms.
Limit Evaluation
Evaluating limits is a fundamental aspect of calculus, aiming to find the behavior of a function as it approaches a particular point. To evaluate the limit in \( \lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x} \), different techniques like factoring, and rationalizing (using the conjugate) are often necessary to simplify expressions and find the actual limit.Here, the conjugate multiplication leads to a simplification that reveals the underlying behavior of the function as \( \Delta x \) approaches zero. After rationalizing, the expression becomes manageable, and we can substitute \( \Delta x = 0 \) directly, providing the solution directly without an undefined or indeterminate form.
Indeterminate Forms
Indeterminate forms in calculus occur when evaluating limits lead to undefined expressions like \( \frac{0}{0} \). These forms are tricky because they do not give enough information about the limit, requiring additional work to resolve.To handle indeterminate forms, techniques such as applying the conjugate method or L'Hôpital's Rule are commonly used. Here, with \( \lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x} \), the expression initially presents an indeterminate form \( \frac{0}{0} \). By multiplying with the conjugate, we eliminate the indeterminate form, simplifying the expression to evaluate the limit correctly. This method is crucial to finding solutions to calculus problems involving complex, undefined initial states.

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Most popular questions from this chapter

Inventory Management The number of units in inventory in a small company is \(N=25\left(2\left\|\frac{t+2}{2}\right\|-t\right), \quad 0 \leq t \leq 12\) where the real number \(t\) is the time in months. (a) Use the greatest integer function of a graphing utility to graph this function, and discuss its continuity. (b) How often must the company replenish its inventory?

Describe the interval(s) on which the function is continuous. $$ f(x)=\frac{1}{2}\|2 x\| $$

Find the limit (if it exists). \(\lim _{\Delta x \rightarrow 0} \frac{2(x+\Delta x)-2 x}{\Delta x}\)

The limit of \(f(x)=(1+x)^{1 / x}\) is a natural base for many business applications, as you will see in Section \(4.2 .\) \(\lim _{x \rightarrow 0}(1+x)^{1 / x}=e \approx 2.718\) (a) Show the reasonableness of this limit by completing the table. \(\begin{array}{|c|c|c|c|c|c|c|}\hline x & {-0.01} & {-0.001} & {-0.0001} & {0} & {0.0001} & {0.001} & {0.01} \\ \hline f(x) & {} & {} & {} & {} & {} \\\ \hline\end{array}\) (b) Use a graphing utility to graph \(f\) and to confirm the answer in part (a). (c) Find the domain and range of the function.

The cost (in dollars) of removing \(p \%\) of the pollutants from the water in a small lake is given by \(C=\frac{25,000 p}{100-p}, \quad 0 \leq p<100\) where \(C\) is the cost and \(p\) is the percent of pollutants. (a) Find the cost of removing \(50 \%\) of the pollutants. (b) What percent of the pollutants can be removed for \(\$ 100,000 ?\) (c) Evaluate \(\lim _{p \rightarrow 100^{-}} C .\) Explain your results.

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