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How do you find the asymptotes for #f(x)= (2x)/(x^2+16)#?

Answer 1

Anna Allen

Since the denominator will never equal zero, there are no vertical asymptotes .

There is one horizontal asymptote:

#""_(xrarr+-oo)^"lim"(2x)/(x^2+16)=0#

Horizontal : #y=0#

Hope that helped

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Answer 2

Anna Allen

To find the asymptotes for ( f(x) = \frac{2x}{x^2 + 16} ), identify the vertical asymptotes by setting the denominator equal to zero and solving for ( x ). In this case, there are no vertical asymptotes. Then, determine the horizontal asymptote by comparing the degrees of the numerator and denominator. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is ( y = 0 ).

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Answer from HIX Tutor

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.