How do you find the asymptotes for #f(x)=(12x)/sqrt(1+x^2) #?
You find the asymptotes by subbing a number into x, in order to find the value which is undefined.
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To find the asymptotes for ( f(x) = \frac{1  2x}{\sqrt{1 + x^2}} ), we examine both the horizontal and vertical behavior of the function.

Vertical asymptotes: The vertical asymptotes occur where the denominator of the function becomes zero, but the numerator does not. In this case, the denominator is ( \sqrt{1 + x^2} ), which equals zero when ( x = \pm i ) (imaginary). Since these are not real values, there are no vertical asymptotes for this function.

Horizontal asymptotes: To find the horizontal asymptote, we examine the behavior of the function as ( x ) approaches positive and negative infinity. We take the limit of the function as ( x ) approaches infinity and negative infinity.
[ \lim_{x \to \infty} \frac{1  2x}{\sqrt{1 + x^2}} ]
[ \lim_{x \to \infty} \frac{2x}{\sqrt{1 + x^2}} = 2 ]
[ \lim_{x \to \infty} \frac{1  2x}{\sqrt{1 + x^2}} ]
[ \lim_{x \to \infty} \frac{2x}{\sqrt{1 + x^2}} = 2 ]
Therefore, the horizontal asymptote of the function is ( y = 2 ).
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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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