Question

Let \( S_n = 1 + 3x + 9x^2 + 27x^3 + \ldots + n \text{ terms} -\frac{1}{3}<x<\frac{1}{3} \). If \( f(x) = S_n \), then \( f(x) \) is discontinuous at the point \( x = \):

• A. 0
• B. \(\frac{1}{3}\)
• C. \(\frac{1}{9}\)
• D. \(-1\)

Hint:

To find discontinuities of a function defined as a limit of a sequence, evaluate the limit within the given interval and check the behavior at the boundaries, especially where the denominator may become zero.

Answer 1

The Correct Option is B

Step 1: Express \( S_n \) in closed form.
The series \( 1 + 3x + 9x^2 + 27x^3 + \ldots + n \text{ terms} \) is a geometric series with first term 1 and common ratio \( 3x \). The sum of \( n \) terms of a geometric series \( a + ar + ar^2 + \ldots + ar^{n-1} \) is: \[ \frac{a(1 - r^n)}{1 - r} \] Here, \( a = 1 \), \( r = 3x \), so the sum of the first \( n \) terms is: \[ 1 + 3x + 9x^2 + \ldots + (3x)^{n-1} = \frac{1 - (3x)^n}{1 - 3x} \] Thus: \[ S_n = \frac{1 - (3x)^n}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} \] So, \( f(x) = S_n \).
Step 2: Analyze the behavior of \( f(x) \) as \( n \to \infty \).
The problem states \( f(x) = S_n \), but in the context of discontinuity and the limit notation \( \lim_{n \to \infty} \), we interpret \( f(x) \) as the limit of the sequence \( S_n \): \[ f(x) = \lim_{n \to \infty} S_n = \lim_{n \to \infty} \left( \frac{1 - (3x)^n}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} \right) \] For \( |3x|<1 \), i.e., \( -\frac{1}{3}<x<\frac{1}{3} \), \( (3x)^n \to 0 \), so: \[ f(x) = \frac{1 - 0}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} = \frac{1}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} \] At \( x = \frac{1}{3} \), \( 3x = 1 \), so \( (3x)^n = 1 \), and the denominator \( 1 - 3x = 0 \), making the first term undefined directly. We need to evaluate the limit. At \( x = -\frac{1}{3} \), \( 3x = -1 \), so \( (3x)^n = (-1)^n \), which oscillates and does not converge, but this is outside the given interval.
Step 3: Check for discontinuities in the interval \( -\frac{1}{3}<x<\frac{1}{3} \).
Within \( -\frac{1}{3}<x<\frac{1}{3} \), \( f(x) = \frac{1}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} \) is continuous, as the denominator \( 1 - 3x \neq 0 \). However, the interval’s endpoints need checking:
At \( x = \frac{1}{3} \), the denominator \( 1 - 3x = 0 \), so \( f(x) \) has a singularity (vertical asymptote), indicating a discontinuity. Compute the limit: \[ \lim_{x \to \frac{1}{3}^-} \left( \frac{1}{1 - 3x} - \frac{1}{3}x + \frac{1}{3} \right) \] As \( x \to \frac{1}{3}^- \), \( 1 - 3x \to 0^+ \), so \( \frac{1}{1 - 3x} \to +\infty \), and the other terms are finite (\(-\frac{1}{3} \cdot \frac{1}{3} + \frac{1}{3} = \frac{2}{9}\)), so \( f(x) \to +\infty \), confirming a discontinuity at \( x = \frac{1}{3} \).
Step 4: Check other options.
At \( x = 0 \): \( f(0) = \frac{1}{1 - 0} - \frac{1}{3}(0) + \frac{1}{3} = 1 + \frac{1}{3} = \frac{4}{3} \), which is finite and continuous.
At \( x = \frac{1}{9} \): \( 3x = \frac{1}{3} \), \( f\left(\frac{1}{9}\right) = \frac{1}{1 - \frac{1}{3}} - \frac{1}{3} \cdot \frac{1}{9} + \frac{1}{3} = \frac{1}{\frac{2}{3}} - \frac{1}{27} + \frac{1}{3} = \frac{3}{2} - \frac{1}{27} + \frac{1}{3} \), which is finite and continuous.
\( x = -1 \) is outside the interval. Thus, the discontinuity occurs at \( x = \frac{1}{3} \). Final Answer: \[ \boxed{2} \]