Locate the discontinuities of the function.
y = 2/(1 + e^(1\/x))
First step, find any vertical asymptotes.
1 + e^(1/x) = 0
e^(1/x) = -1
1/x = ln(-1)
Obviously since this is impossible, there are no vertical asymptotes.
Since we have a fraction in the function (1/x), and the x is in the denominator, we can conclude that f(x) does not exist at 0, since it is undefined. We still have to make sure the limit does not exist, either though. lim(x->0+)f(x) = 0, lim(x->0-)f(x) = 2. Therefore, lim(x->0)f(x) does not exist. :)
That's it!
To locate the discontinuities of the function y = 2/(1 + e^(1\/x)), we need to look for values of x where the function is undefined or not continuous.
First, let's consider the denominator of the function, which is (1 + e^(1/x)). The exponential term e^(1/x) is defined for any real value of x, since the base of the exponential function is positive.
However, we need to pay attention to the term 1/x, as x approaches zero. As x approaches zero from the left side (negative side), the term 1/x approaches negative infinity, and as x approaches zero from the right side (positive side), the term 1/x approaches positive infinity.
Therefore, at x = 0, the term 1/x is undefined because we cannot divide by zero.
Now, let's analyze the whole function. As long as the denominator is non-zero, the function y = 2/(1 + e^(1/x)) is continuous.
The only discontinuity in this case is at x = 0, where the denominator becomes zero.
Hence, the function y = 2/(1 + e^(1\/x)) is discontinuous or undefined at x = 0.