True or false. Explain.
The line x=1 is a vertical asymptote of f(x)=(x^2+x-2)/(x^2-1)
well, if x = .99 I get 1.50
if x = .999 then 1.500
that is because
numerator = (x-1)(x+2)
denominator = (x-1)(x+1)
so what we really have is
(x+2)/(x+1) which is 1.5 at x = 1
It is true that the original function is undefined at x = 1
False.
To determine if the line x=1 is a vertical asymptote of the function f(x)=(x^2+x-2)/(x^2-1), we need to check the behavior of the function as x approaches 1 from both the left and right sides.
1. Calculate the limit as x approaches 1 from the left side:
- Evaluate the function for values of x slightly less than 1, such as 0.9, 0.99, 0.999, and so on.
- Calculate the resulting y-values of the function for these x-values.
- If the resulting y-values become extremely large (positive or negative) or tend towards positive or negative infinity as x approaches 1 from the left side, then x=1 is a vertical asymptote.
2. Calculate the limit as x approaches 1 from the right side:
- Evaluate the function for values of x slightly greater than 1, such as 1.1, 1.01, 1.001, and so on.
- Calculate the resulting y-values of the function for these x-values.
- If the resulting y-values become extremely large (positive or negative) or tend towards positive or negative infinity as x approaches 1 from the right side, then x=1 is a vertical asymptote.
In the case of f(x)=(x^2+x-2)/(x^2-1), if you evaluate the function for x-values close to 1 from both sides, you will find that the y-values do not become extremely large or tend towards infinity. Therefore, the line x=1 is not a vertical asymptote of the function.