## anonymous one year ago Horizontal asymptote question. Confused about values.

1. anonymous

2. anonymous

I can't use L'hopital's rule on this. How else can I take this limit?

3. triciaal

@jim_thompson5910

4. anonymous

Nvm I got it.

5. phi

It took me forever to notice the problem $\lim_{x\rightarrow -\infty} \frac{x+2}{\sqrt{x^2+1}-x}$ you can divide top and bottom by x to get $\lim_{x\rightarrow -\infty} \frac{1+\frac{2}{x}}{\frac{\sqrt{x^2+1}}{x}-1}$ and we must be careful with the square root term when x is negative $\frac{\sqrt{x^2+1}}{x}$ the top is always positive, so when we divide by a negative number, the results is negative: $- \frac{ \sqrt{x^2+1}}{\sqrt{x^2}} = - \sqrt{1+\frac{1}{x^2} }$ now we can take the limit $\lim_{x\rightarrow -\infty} \frac{1+\frac{2}{x}}{ - \sqrt{1+\frac{1}{x^2} }-1} = - \frac{1}{2}$