Chapter 12: Problem 41
The values of \(p\) and \(q\) for which the function \(f(x)= \begin{cases}\frac{\sin (p+1) x+\sin x}{x}, & x<0 \\ q, & x=0 \\\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}}, & x=0\end{cases}\) is continuous for all \(x\) in \(R\), are (A) \(p=\frac{1}{2}, q=\frac{3}{2}\) (B) \(p=\frac{1}{2}, q=-\frac{3}{2}\) (C) \(p=\frac{5}{2}, q=\frac{1}{2}\) (D) \(p=-\frac{3}{2}, q=\frac{1}{2}\)
Short Answer
Step by step solution
Understanding the Problem
Continuity at x = 0
Calculating Left-Hand Limit
Calculating Right-Hand Limit
Setting Continuity Condition at x=0
Solving for p and q
Conclusion - Validate
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Limits and Continuity
- The function must be defined at that point, meaning there's a value for the function at this specific point.
- The limit of the function, as it approaches the point from the left, must equal the limit as it approaches from the right.
- These two limits must equal the value of the function at the point.
L'Hospital's Rule
In the exercise, L'Hospital's Rule was applied in finding the limit of the left-hand portion of the piecewise function as \(x\) approaches 0 from the negative side, specifically:\[\lim_{x \to 0^-} \frac{\sin((p+1)x) + \sin(x)}{x}\]By differentiating the sine functions' sums and the \(x\) in the denominator separately, the expression resolved from an indeterminate form to a solvable equation \(p + 2\). Thus, mastering L'Hospital's Rule aids in breaking down limits that initially seem unsolvable into simpler computations.
Piecewise Functions
- The function is continuous at \(x=0\) if the expressions before and after the partition point render the same result at that specific point.
- The continuity condition bridges these definitions together by ensuring the consistency of the limit behavior on either side of the partition.