How do you find the horizontal asymptote for #f(x) = (3x^2 + 2x  1) / (x + 1)#?
You evaluate the limits of the function as x approaches infinities (
There are no horizontal asymptotes in this function.
Possible horizontals
Solution
Solving the numerator:
So the numerator can be written as follows:
The two limits are now easier to find:
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To find the horizontal asymptote of the function ( f(x) = \frac{3x^2 + 2x  1}{x + 1} ), you can look at the behavior of the function as ( x ) approaches positive infinity or negative infinity.

As ( x ) approaches positive infinity: [ f(x) = \frac{3x^2 + 2x  1}{x + 1} ] [ \lim_{{x \to +\infty}} f(x) = \lim_{{x \to +\infty}} \frac{3x^2 + 2x  1}{x + 1} ] [ \lim_{{x \to +\infty}} f(x) = \lim_{{x \to +\infty}} \frac{3x^2}{x} = \lim_{{x \to +\infty}} 3x = +\infty ]

As ( x ) approaches negative infinity: [ f(x) = \frac{3x^2 + 2x  1}{x + 1} ] [ \lim_{{x \to \infty}} f(x) = \lim_{{x \to \infty}} \frac{3x^2 + 2x  1}{x + 1} ] [ \lim_{{x \to \infty}} f(x) = \lim_{{x \to \infty}} \frac{3x^2}{x} = \lim_{{x \to \infty}} 3x = \infty ]
Since the limit as ( x ) approaches both positive and negative infinity is either positive or negative infinity, there is no horizontal asymptote for this function.
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When evaluating a onesided 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 onesided 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 onesided 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 onesided 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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