Question

Horizontal asymptote question. Confused about values.

Answer 1

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

Answer 2

@jim_thompson5910

Answer 3

Nvm I got it.

Answer 4

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
\[ \lim_{x\rightarrow -\infty} \frac{1+\frac{2}{x}}{ - \sqrt{1+\frac{1}{x^2} }-1}
= - \frac{1}{2}
\]