Problem 17
Question
\(\lim _{x \rightarrow a^{-}} \frac{\sqrt{x-b}-\sqrt{a-b}}{\left(x^{2}-a^{2}\right)},(a>b)\) (a) \(\frac{1}{4 a}\) (b) \(\frac{1}{a \sqrt{a-b}}\) (c) \(\frac{1}{2 a \sqrt{a-b}}\) (d) \(\frac{1}{4 a \sqrt{a-b}}\)
Step-by-Step Solution
Verified Answer
The correct answer is (d): \(\frac{1}{4a\sqrt{a-b}}\).
1Step 1: Identify the Type of Limit
The limit problem involves a difference of square roots in the numerator and a difference of squares in the denominator. This suggests the use of algebraic manipulation to simplify.
2Step 2: Rationalize the Numerator
Multiply and divide the expression by the conjugate of the numerator: \(\frac{\sqrt{x-b}-\sqrt{a-b}}{x^2-a^2} \times \frac{\sqrt{x-b}+\sqrt{a-b}}{\sqrt{x-b}+\sqrt{a-b}}\). This gives: \[\frac{(\sqrt{x-b})^2-(\sqrt{a-b})^2}{(x^2-a^2)(\sqrt{x-b}+\sqrt{a-b})}\].
3Step 3: Simplify the Numerator
Use the identity \((A-B)(A+B) = A^2 - B^2\). This simplifies the numerator: \((x-b) - (a-b) = x-a\).
4Step 4: Simplify the Denominator
The expression is now \(\frac{x-a}{(x^2-a^2)(\sqrt{x-b}+\sqrt{a-b})}\). Factor \(x^2-a^2\) as \((x-a)(x+a)\).
5Step 5: Cancel Common Factors
Cancel the factor \(x-a\) from numerator and denominator. The expression simplifies to \(\frac{1}{(x+a)(\sqrt{x-b}+\sqrt{a-b})}\).
6Step 6: Evaluate the Limit as \(x \to a^-\)
Substitute \(x = a\) in the simplified expression: \[\frac{1}{(a+a)(\sqrt{a-b}+\sqrt{a-b})} = \frac{1}{2a\times2\sqrt{a-b}} = \frac{1}{4a\sqrt{a-b}}\].
7Step 7: Choose the Correct Answer
The correct answer corresponds to option (d): \(\frac{1}{4a\sqrt{a-b}}\).
Key Concepts
Rationalization TechniqueL'Hôpital's RuleAlgebraic Simplification in Calculus
Rationalization Technique
The rationalization technique is a special method in calculus used to simplify expressions involving roots, such as square roots, cube roots, etc. It is particularly useful in limit problems where the expression includes roots in the numerator or denominator. By rationalizing, we aim to remove the roots so that simplifications become easier and clearer.
In the original exercise, the expression in the numerator is \( \sqrt{x-b} - \sqrt{a-b} \), which involves a difference of square roots. To rationalize this expression, we multiply and divide by the conjugate, \( \sqrt{x-b} + \sqrt{a-b} \). The conjugate changes the expression into a differencing of squares, eliminating the roots:
In the original exercise, the expression in the numerator is \( \sqrt{x-b} - \sqrt{a-b} \), which involves a difference of square roots. To rationalize this expression, we multiply and divide by the conjugate, \( \sqrt{x-b} + \sqrt{a-b} \). The conjugate changes the expression into a differencing of squares, eliminating the roots:
- Multiply: \( (\sqrt{x-b})^2 - (\sqrt{a-b})^2 \)
- Simplify: \( (x-b) - (a-b) = x-a \)
L'Hôpital's Rule
L'Hôpital's Rule is valuable when resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). Although not explicitly used in our example, it's crucial in similar problems where these forms appear after initial simplifications.
To apply L'Hôpital's Rule, you must first evaluate the function and confirm it results in an indeterminate form. If permissible, you then differentiate the numerator and the denominator separately and re-evaluate the limit using the resulting derivatives. Repeat this process until the indeterminate form resolves.
To apply L'Hôpital's Rule, you must first evaluate the function and confirm it results in an indeterminate form. If permissible, you then differentiate the numerator and the denominator separately and re-evaluate the limit using the resulting derivatives. Repeat this process until the indeterminate form resolves.
- Confirm the form: \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \)
- Differentiate numerator and denominator
- Evaluate or repeat if necessary
Algebraic Simplification in Calculus
Algebraic simplification is a cornerstone in solving calculus problems, especially limits. It involves manipulating expressions to reveal simpler forms, making evaluation more straightforward. In our original problem, we see multiple steps of algebraic simplification to ease the calculation of the limit.
After rationalization, the initial complex fraction becomes manageable. By identifying common formulas such as \( (x^2 - a^2) = (x-a)(x+a) \) and simplifying accordingly:
After rationalization, the initial complex fraction becomes manageable. By identifying common formulas such as \( (x^2 - a^2) = (x-a)(x+a) \) and simplifying accordingly:
- Numerator: From \( (x-b)-(a-b) = x-a \)
- Denominator: Factor \( x^2-a^2 \) as \( (x-a)(x+a) \)
- Cancel common terms: Remove \( x-a \) from both top and bottom
Other exercises in this chapter
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Let \([x]\) denotes the greatest integer function. Let \(g(x)=\frac{\sin \frac{\pi}{4}[x]}{[x]}\), then \(g\) is such that (a) it is continuous at \(x=\frac{3}{
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View solution Problem 18
\(\lim _{n \rightarrow \infty}\left(\sin ^{n} 1+\cos ^{n} 1\right)^{n}\) is equal to (a) \(\cot 1\) (b) \(\tan 1\) (c) \(\cos 1\) (d) \(\sin 1\)
View solution Problem 19
If \(\lim _{x \rightarrow \infty} 4 x\left(\frac{\pi}{4}-\tan ^{-1} \frac{x+1}{x+2}\right)=y^{2}+4 y+5\), then \(y\) can be equal to (a) 1 (b) \(-1\) (c) \(-4\)
View solution