How do you find the horizontal asymptote for #f(x)=(1-2x)/sqrt(1+x^2)#?

Question

Sadie Ackerman · Accepted Answer

The horizontal asymptotes are #y=-2# if #x\to\infty# and #y=2# if #x\to-\infty#

Horizontal asymptotes, if exist, are given by

#lim_{x\to\pm\infty} f(x)#

So, to compute your limit, observe that

#\frac{1-2x}{sqrt(1+x^2)}=\frac{x(-2+1/x)}{sqrt(x^2(1+1/x^2))} =\frac{x(-2+1/x)}{abs(x)sqrt((1+1/x^2))}#

where the last step is due to the fact that #sqrt(x^2)=abs(x)#

Now, when #x\to\pm\infty#, we have that both #1/x# and #1/x^2# tend to zero. On the other hand, #abs(x)=-x# if #x\to-\infty#, and #abs(x)=x# if #x\to\infty#. So, the limit becomes

#lim_{x\to\pm\infty} \frac{x(-2+cancel(1/x))}{abs(x)sqrt((1+cancel(1/x^2)))} = \pm1 * -2/sqrt(1) = -(\pm2)#

This means that horizontal asymptotes are #y=-2# if #x\to\infty# and #y=2# if #x\to-\infty#

Chloe Alexander · Answer

undefined To find the horizontal asymptote for $ f(x) = \frac{1 - 2x}{\sqrt{1 + x^2}} $, you examine the behavior of the function as $ x $ approaches positive and negative infinity. Since the degree of the numerator is equal to the degree of the denominator, you divide the leading coefficients of both to find the horizontal asymptote. In this case, the leading coefficient of the numerator is -2, and the leading coefficient of the denominator is 1. Therefore, the horizontal asymptote is $ y = -2 $.