anonymous
  • anonymous
Horizontal asymptote question. Confused about values.
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
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anonymous
  • anonymous
I can't use L'hopital's rule on this. How else can I take this limit?
triciaal
  • triciaal

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anonymous
  • anonymous
Nvm I got it.
phi
  • 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 \[ \lim_{x\rightarrow -\infty} \frac{1+\frac{2}{x}}{ - \sqrt{1+\frac{1}{x^2} }-1} = - \frac{1}{2} \]

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