Determine the interval(s) on which the following function is continuous. Then analyze the given limits.
\[
f(x)=\frac{2 e^{x}}{1-e^{x}} ; \quad \lim _{x \rightarrow 0^{-}} f(x) ; \quad \lim _{x \rightarrow 0^{+}} f(x)
\]
Answer
Final Answer: The function is continuous for all real numbers except \(\boxed{x = 0}\). As x approaches 0 from the left, the limit of the function is \(\boxed{\infty}\), and as x approaches 0 from the right, the limit of the function is \(\boxed{-\infty}\).
Step 1 ：The function is a rational function, and rational functions are continuous everywhere except where the denominator is zero. So, we need to find the values of x for which the denominator of the function, \(1 - e^x\), is zero.
Step 2 ：The denominator of the function is zero at \(x = 0\), so the function is not continuous at \(x = 0\).
Step 3 ：For the limits, we need to find the value of the function as x approaches 0 from the left and from the right.
Step 4 ：As x approaches 0 from the left, the function approaches positive infinity.
Step 5 ：As x approaches 0 from the right, the function approaches negative infinity.
Step 6 ：Final Answer: The function is continuous for all real numbers except \(\boxed{x = 0}\). As x approaches 0 from the left, the limit of the function is \(\boxed{\infty}\), and as x approaches 0 from the right, the limit of the function is \(\boxed{-\infty}\).
