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Sagot :
To determine the interval(s) on which the function is continuous and to analyze the limits as [tex]\( x \)[/tex] approaches 0 from both sides, follow these detailed steps.
We are given the function:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \][/tex]
### Step 1: Identify Points of Discontinuity
To find where the function is continuous, we need to examine the denominator [tex]\( 1 - e^x \)[/tex]. The function will be undefined where the denominator is zero, as division by zero is undefined.
Set the denominator equal to zero:
[tex]\[ 1 - e^x = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ e^x = 1 \][/tex]
Recall that the exponential function [tex]\( e^x \)[/tex] equals 1 when [tex]\( x = 0 \)[/tex]. Therefore, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 0 \)[/tex]. Hence, the function may only be continuous in the intervals where [tex]\( x \neq 0 \)[/tex].
Thus, the function is continuous on:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
### Step 2: Analyze the Limit as [tex]\( x \to 0^- \)[/tex]
To find the limit as [tex]\( x \)[/tex] approaches 0 from the left, we need to determine [tex]\( \lim_{x \to 0^-} f(x) \)[/tex]. Observe the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 from negative values:
As [tex]\( x \to 0^- \)[/tex], [tex]\( e^x \)[/tex] approaches 1 from values less than 1 (since the exponential function is always positive and increases toward 1). This means the denominator [tex]\( 1 - e^x \)[/tex] is a small positive number, becoming closer to 0.
Therefore:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \text{ becomes very large negatively} \][/tex]
as the numerator [tex]\( 5 e^x \)[/tex] (close to 5) remains finite and the denominator approaches 0 positively. Hence:
[tex]\[ \lim_{x \to 0^-} f(x) = -\infty \][/tex]
### Step 3: Analyze the Limit as [tex]\( x \to 0^+ \)[/tex]
To find the limit as [tex]\( x \)[/tex] approaches 0 from the right, we need to determine [tex]\( \lim_{x \to 0^+} f(x) \)[/tex]. Observe the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 from positive values:
As [tex]\( x \to 0^+ \)[/tex], [tex]\( e^x \)[/tex] approaches 1 from values greater than 1. In this context, the denominator [tex]\( 1 - e^x \)[/tex] is a small negative number, becoming closer to 0.
Therefore:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \text{ becomes very large positively} \][/tex]
as the numerator [tex]\( 5 e^x \)[/tex] remains finite and the denominator approaches 0 negatively. Hence:
[tex]\[ \lim_{x \to 0^+} f(x) = +\infty \][/tex]
### Summary
In conclusion, the function [tex]\( f(x) = \frac{5 e^x}{1 - e^x} \)[/tex] is continuous on:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
The limits as [tex]\( x \)[/tex] approaches 0 from both sides are:
[tex]\[ \lim_{x \to 0^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 0^+} f(x) = +\infty \][/tex]
We are given the function:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \][/tex]
### Step 1: Identify Points of Discontinuity
To find where the function is continuous, we need to examine the denominator [tex]\( 1 - e^x \)[/tex]. The function will be undefined where the denominator is zero, as division by zero is undefined.
Set the denominator equal to zero:
[tex]\[ 1 - e^x = 0 \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ e^x = 1 \][/tex]
Recall that the exponential function [tex]\( e^x \)[/tex] equals 1 when [tex]\( x = 0 \)[/tex]. Therefore, the function [tex]\( f(x) \)[/tex] is undefined at [tex]\( x = 0 \)[/tex]. Hence, the function may only be continuous in the intervals where [tex]\( x \neq 0 \)[/tex].
Thus, the function is continuous on:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
### Step 2: Analyze the Limit as [tex]\( x \to 0^- \)[/tex]
To find the limit as [tex]\( x \)[/tex] approaches 0 from the left, we need to determine [tex]\( \lim_{x \to 0^-} f(x) \)[/tex]. Observe the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 from negative values:
As [tex]\( x \to 0^- \)[/tex], [tex]\( e^x \)[/tex] approaches 1 from values less than 1 (since the exponential function is always positive and increases toward 1). This means the denominator [tex]\( 1 - e^x \)[/tex] is a small positive number, becoming closer to 0.
Therefore:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \text{ becomes very large negatively} \][/tex]
as the numerator [tex]\( 5 e^x \)[/tex] (close to 5) remains finite and the denominator approaches 0 positively. Hence:
[tex]\[ \lim_{x \to 0^-} f(x) = -\infty \][/tex]
### Step 3: Analyze the Limit as [tex]\( x \to 0^+ \)[/tex]
To find the limit as [tex]\( x \)[/tex] approaches 0 from the right, we need to determine [tex]\( \lim_{x \to 0^+} f(x) \)[/tex]. Observe the behavior of [tex]\( f(x) \)[/tex] as [tex]\( x \)[/tex] approaches 0 from positive values:
As [tex]\( x \to 0^+ \)[/tex], [tex]\( e^x \)[/tex] approaches 1 from values greater than 1. In this context, the denominator [tex]\( 1 - e^x \)[/tex] is a small negative number, becoming closer to 0.
Therefore:
[tex]\[ f(x) = \frac{5 e^x}{1 - e^x} \text{ becomes very large positively} \][/tex]
as the numerator [tex]\( 5 e^x \)[/tex] remains finite and the denominator approaches 0 negatively. Hence:
[tex]\[ \lim_{x \to 0^+} f(x) = +\infty \][/tex]
### Summary
In conclusion, the function [tex]\( f(x) = \frac{5 e^x}{1 - e^x} \)[/tex] is continuous on:
[tex]\[ (-\infty, 0) \cup (0, \infty) \][/tex]
The limits as [tex]\( x \)[/tex] approaches 0 from both sides are:
[tex]\[ \lim_{x \to 0^-} f(x) = -\infty \quad \text{and} \quad \lim_{x \to 0^+} f(x) = +\infty \][/tex]
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