How can I find the horizontal asymptote of:

y=e^x/(2e^x+5)

i know you have to do the limit as x approaches infinity... but that is about all i know...

mulitply the numberator and denominator by e^-x

y= e^0/(2e^0 + 5e^-x)

y= 1/(2 + 5e^-x)

now, take lim as x >> int. Notice the e^-x goes to zero.

y= 1/2 as x approaches zero.

To find the horizontal asymptote of a function, you need to take the limit as x approaches infinity (or negative infinity) for the given function. Let's go step by step to find the horizontal asymptote for the function y = e^x / (2e^x + 5).

1. Start by multiplying both the numerator and denominator by e^(-x). This is done to simplify the expression and get rid of the exponential term in the denominator.

y = (e^x * e^(-x)) / ((2e^x * e^(-x)) + 5e^(-x))

Simplifying this expression gives us:

y = 1 / (2 + 5e^(-x))

2. Now, let's take the limit as x approaches infinity. In this case, we're interested in knowing the behavior of the function as x gets very large.

lim(x→∞) (1 / (2 + 5e^(-x)))

3. Notice that as x approaches infinity, the term e^(-x) becomes very small (approaching zero). This means that the denominator (2 + 5e^(-x)) will be dominated by the constant term 2.

So, we can disregard the 5e^(-x) term and keep only the 2.

lim(x→∞) (1 / 2)

4. Therefore, as x approaches infinity, the function y approaches the value 1/2. This means that the horizontal asymptote of the given function is y = 1/2.

In summary, the horizontal asymptote of y = e^x / (2e^x + 5) is y = 1/2.