# How do you find the vertical, horizontal or slant asymptotes for #f(x)= (3e^(x))/(2-2e^(x))#?

Vertical asymptote is

two horizontal asymptotes

Vertical asymptote is given by putting denominator equal to zero.

graph{3e^x/(2-2e^x) [-10, 10, -5, 5]}

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To find the vertical asymptotes of the function ( f(x) = \frac{3e^x}{2 - 2e^x} ), we need to identify the values of ( x ) for which the denominator equals zero. Setting the denominator equal to zero, we solve for ( x ):

[ 2 - 2e^x = 0 ]

[ 2 = 2e^x ]

[ 1 = e^x ]

[ x = \ln(1) ]

Since ( \ln(1) = 0 ), there is no vertical asymptote for this function.

To find the horizontal asymptote, we examine the behavior of the function as ( x ) approaches positive or negative infinity.

As ( x ) approaches positive infinity, both the numerator and the denominator grow without bound. Therefore, we can apply L'Hôpital's Rule:

[ \lim_{x \to \infty} \frac{3e^x}{2 - 2e^x} = \lim_{x \to \infty} \frac{3e^x}{-2e^x} = -\frac{3}{2} ]

So, the horizontal asymptote is ( y = -\frac{3}{2} ).

For the slant asymptote, we can use polynomial long division to divide the numerator by the denominator. However, in this case, the degree of the numerator (1) is less than the degree of the denominator (0). Therefore, there is no slant asymptote for this function.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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