Chapter 12: Problem 115
Assertion: The function \(f(x)=\) \(\lim _{n \rightarrow \infty} \frac{\cos \pi x-x^{2 n} \sin (x-1)}{1+x^{2 n+1}-x^{2 n}}\) is discontinuous at \(x=\pm 1\) Reason: \(f(x)=\left\\{\begin{array}{cl}\frac{\cos \pi x}{1+x}, & |x|<1 \\\ -1+\sin 2, & x=-1 \\ -1, & x=1 \\ \frac{-\sin (x-1)}{x-1}, & |x|>1\end{array}\right.\)
Short Answer
Step by step solution
Understanding the Limit
Simplifying for \(|x| < 1\)
Analyzing at \(x = 1\) and \(x = -1\)
Simplifying for \(|x| > 1\)
Evaluating Continuity At \(x = \pm 1\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Continuity
- The function is defined at the point.
- The limit of the function as it approaches the point exists.
- The limit of the function equals the function's value at that point.
Discontinuity
- Jump Discontinuity: The function has different limits approaching from the left and right.
- Infinite Discontinuity: The limit does not exist due to unbounded behavior at the point.
- Removable Discontinuity: The limit exists, but the function value at the point is not defined or not equal to the limit.
Function behavior
- For \( |x| < 1 \), the function simplifies, indicating a predictable behavior dictated by \( \cos \pi x \).
- For \( |x| > 1 \), the dominance of \( x^{2n} \) suggests different control, resulting in a function that looks like \( \frac{-\sin(x-1)}{x-1} \).
Limit evaluation
- For \( |x| < 1 \), as \( n \to \infty \), the function converges to \( \cos \pi x \) simplifying the overall expression.
- For \( x = 1 \) and \( x = -1 \), the need is to precisely define the behavior because terms in both the numerator and denominator greatly influence the function's outcome.
- For \( |x| > 1 \), \( x^{2n} \to \infty \) leading to significant simplifications.