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

1 Answer
Dec 6, 2015

Answer:

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

Explanation:

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#