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Determine the interval(s) on which the following function is continuous. Then analyze the given limits. \( f(x)=\frac{e^{x}}{1-e^{6 x}} ; \lim _{x \rightarrow 0^{-}} f(x) ; \lim _{x \rightarrow 0^{+}} f(x) \) The function is continuous on the interval(s) (Simplify your answer. Type your answer in interval notation. Use a comma to separate answers as needed.

Ask by Ellis Kelley. in the United States
Feb 02,2025

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Answer

The function \( f(x) = \frac{e^{x}}{1 - e^{6x}} \) is continuous on the intervals \( (-\infty, 0) \) and \( (0, \infty) \). For the limits: \[ \lim_{x \rightarrow 0^{-}} f(x) = +\infty \] \[ \lim_{x \rightarrow 0^{+}} f(x) = -\infty \]

Solution

To analyze the continuity of the function \( f(x) = \frac{e^{x}}{1 - e^{6x}} \) and evaluate the given limits, follow these steps: ### 1. Determine the Intervals of Continuity The function \( f(x) = \frac{e^{x}}{1 - e^{6x}} \) is continuous wherever its denominator is not zero and both the numerator and denominator are defined. - **Denominator Analysis:** \[ 1 - e^{6x} = 0 \quad \Rightarrow \quad e^{6x} = 1 \quad \Rightarrow \quad 6x = 0 \quad \Rightarrow \quad x = 0 \] The denominator is zero only at \( x = 0 \), which is the point of discontinuity. - **Intervals of Continuity:** Therefore, the function is continuous on all real numbers except at \( x = 0 \). In interval notation, the intervals of continuity are: \[ (-\infty, 0), \quad (0, \infty) \] ### 2. Evaluate the Limits #### a. \( \lim_{x \rightarrow 0^{-}} f(x) \) As \( x \) approaches \( 0 \) from the negative side: - \( e^{x} \) approaches \( 1 \). - \( e^{6x} \) approaches \( 1 \) **from below** since \( 6x \) is negative and approaching \( 0 \). Thus: \[ 1 - e^{6x} \rightarrow 0^{+} \] \[ f(x) = \frac{e^{x}}{1 - e^{6x}} \rightarrow \frac{1}{0^{+}} = +\infty \] **Conclusion:** \[ \lim_{x \rightarrow 0^{-}} \frac{e^{x}}{1 - e^{6x}} = +\infty \] #### b. \( \lim_{x \rightarrow 0^{+}} f(x) \) As \( x \) approaches \( 0 \) from the positive side: - \( e^{x} \) approaches \( 1 \). - \( e^{6x} \) approaches \( 1 \) **from above** since \( 6x \) is positive and approaching \( 0 \). Thus: \[ 1 - e^{6x} \rightarrow 0^{-} \] \[ f(x) = \frac{e^{x}}{1 - e^{6x}} \rightarrow \frac{1}{0^{-}} = -\infty \] **Conclusion:** \[ \lim_{x \rightarrow 0^{+}} \frac{e^{x}}{1 - e^{6x}} = -\infty \] ### Summary - **Intervals of Continuity:** \( (-\infty, 0), \ (0, \infty) \) - **Limits:** \[ \lim_{x \rightarrow 0^{-}} f(x) = +\infty \] \[ \lim_{x \rightarrow 0^{+}} f(x) = -\infty \]

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Beyond the Answer

The function \( f(x) = \frac{e^{x}}{1 - e^{6x}} \) is continuous wherever the denominator is not equal to zero. The expression \( 1 - e^{6x} = 0 \) leads us to find the points of discontinuity by solving for \( x \): \( e^{6x} = 1 \) gives \( 6x = 0 \), or \( x = 0 \). Thus, the function is continuous on the intervals \( (-\infty, 0) \) and \( (0, \infty) \). For the limits, as \( x \rightarrow 0^{-} \), the denominator approaches \( 1 - 1 = 0 \) from the negative side, thus \( f(x) \rightarrow -\infty \). As \( x \rightarrow 0^{+} \), the denominator similarly approaches zero from the positive side, leading to \( f(x) \rightarrow +\infty \).

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