Question

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

Answer 1

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`