Simple limit question

[jwpaine]

Let f(x) = \(\displaystyle \L \frac{\sqrt{x^{6} + 1}}{x^{3} - 1}\)

determine \(\displaystyle \L \lim_{x\to +\infty}f(x)\)

so I focused on the highest degree terms:

\(\displaystyle \L \lim_{x\to +\infty} \frac{\sqrt{x^{6} + 1}}{x^{3} - 1}\)

=\(\displaystyle \L \lim_{x\to +\infty}\frac{\sqrt{x^{6}}}{x^{3}}\)

=\(\displaystyle \L \lim_{x\to +\infty}\frac{x^{3}}{x^{3}} = 1\)


Now question #1: How would I show for \(\displaystyle \L \lim_{x\to -\infty}f(x)\) without just saying -1?



question #2: am I doing this correctly? I should cancel out as much as possible before pluging and chugging... but I don't think I can reduce this fraction, any....

determine \(\displaystyle \L \lim_{x\to 1+}f(x)\)

\(\displaystyle \L \lim_{x\to 1+} \frac{\sqrt{x^{6} + 1}}{x^{3} - 1}\)

= \(\displaystyle \L \lim_{x\to 1+} \frac{\sqrt{(1+)^{6} + 1}}{(1+)^{3} - 1}\)

= \(\displaystyle \L \lim_{x\to 1+} \frac{\sqrt{2+}}{0+} = +\infty\)

Thanks,
John

 

[pka]

This answers #1.
\(\displaystyle \L \frac{{\sqrt {x^6 + 1} }}{{x^3 - 1}} = \frac{{\sqrt {x^6 + 1} }}{{x^3 - 1}}\frac{{\left| {\frac{1}{{x^3 }}} \right|}}{{\left| {\frac{1}{{x^3 }}} \right|}} = \frac{{\sqrt {1 + \frac{1}{{x^6 }}} }}{{x^3 \left| {\frac{1}{{x^3 }}} \right| - \left| {\frac{1}{{x^3 }}} \right|}}\)

 

[jwpaine]

Thank you very much, Pka... that helped me understand a lot better... because that means that the denominator is going to be (-1 - 0) or (1 - 0)

John.