Problem 57
Question
find the limit $$ \lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x} $$
Step-by-Step Solution
Verified Answer
The limit of the function as \(\Delta x\) approaches 0 is \(\frac{1}{2\sqrt{x+2}}\).
1Step 1: Recognize the Problem
The original expression is of the form 0/0 when \(\Delta x\) approaches 0, which is called an indeterminate form. One common method to resolve this is multiplying the expression by its conjugate.
2Step 2: Multiply the Expression by Its Conjugate
Thus, multiply by the conjugate of the numerator over itself: \[\lim _{\Delta x \rightarrow 0} \frac{\sqrt{x+2+\Delta x}-\sqrt{x+2}}{\Delta x} \cdot \frac{\sqrt{x+2 + \Delta x} + \sqrt{x+2}}{\sqrt{x+2 + \Delta x} + \sqrt{x+2}}\]This results in: \[\lim _{\Delta x \rightarrow 0} \frac{(x+2+\Delta x)-(x+2)}{\Delta x(\sqrt{x+2 + \Delta x} + \sqrt{x+2})}\] After comparisons, the numerator becomes \(\Delta x\) and will cancel out with the denominator. The resulting expression is: \[\lim _{\Delta x \rightarrow 0} \frac{1}{\sqrt{x+2 + \Delta x} + \sqrt{x+2}}\].
3Step 3: Evaluate the Limit
Now \(\Delta x\) can approach zero: \(\frac{1}{\sqrt{x+2} + \sqrt{x+2}}\], which simplifies down to \(\frac{1}{2\sqrt{x+2}}\). Therefore, the limit of the function as \(\Delta x\) approaches zero is \(\frac{1}{2\sqrt{x+2}}\).
Key Concepts
Indeterminate formConjugate multiplicationSimplifying expressions
Indeterminate form
When evaluating limits, you might encounter an expression that morphs into a form like \( \frac{0}{0} \) as the variable approaches a certain value. This is known as an indeterminate form. These forms do not provide a clear value just by substituting the limit into the expression directly. Therefore, some work is needed to solve them by transforming or simplifying the expression first. Encountering an indeterminate form means the expression requires a more in-depth analysis to work through the limit accurately. This is where using methods like conjugate multiplication or simplifying expressions come into play. By resolving indeterminate forms, we can accurately determine the limit value of the original function without leaving it undefined. In limit problems, recognizing an indeterminate form is the critical first step. It guides students on choosing the appropriate method to apply, such as conjugate multiplication, to proceed with the solution.
Conjugate multiplication
Conjugate multiplication is a powerful technique for simplifying expressions, especially when you deal with square roots in the numerator or denominator. The conjugate of a binomial is essentially the same expression with the opposite sign between the two terms.For example, if you have the expression \( \sqrt{x+2+\Delta x} - \sqrt{x+2} \), its conjugate would be \( \sqrt{x+2+\Delta x} + \sqrt{x+2} \). By multiplying the original expression by this conjugate over itself, you aren't changing the value of the expression but simply transforming it to a form that is easier to work with.
- This multiplication results in the disappearance of square roots in the numerator, as it employs the difference of squares formula: \((a-b)(a+b) = a^2 - b^2\).
- This simplification often eliminates the indeterminate form (like \( \frac{0}{0} \)) because the original problematic terms usually cancel each other out.
Simplifying expressions
Simplifying expressions is crucial when working with complicated limit problems. After using conjugate multiplication, a new opportunity is there to simplify by canceling terms or substituting the approaching value directly into the expression.In the given scenario, multiplying by the conjugate results in a far simpler expression where the problematic \( \Delta x \) cancels out with itself. The expression is then transformed into \[ \lim _{\Delta x \rightarrow 0} \frac{1}{\sqrt{x+2 + \Delta x} + \sqrt{x+2}} \]. From here:
- Set \( \Delta x \) to approach zero: the expression becomes \( \frac{1}{\sqrt{x+2} + \sqrt{x+2}} \).
- Further simplify by merging terms: \( \frac{1}{2\sqrt{x+2}} \).
Other exercises in this chapter
Problem 57
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