Question

Need help with this. Calc final in 4 hours!
lim{x to infinity} (cuberoot{(8^x+2^x+1)/(1-3.2^{3x})})
The answer comes out as -1/cuberoot{3}. How??

Answer 1

@ParthKohli

Answer 2

Are you familiar with l'hoptal's rule?

Answer 3

yeah

Answer 4

\(\Large \lim_{a \rightarrow b}[\frac{f'(x)}{g'(x)}]\)

Answer 5

Did you use l'hoptal's rule?

Answer 6

alternatively try dividing \(8^x\) top and bottom inside the radical

Answer 7

\[\lim\limits_{x\to\infty}~\sqrt[3]{\dfrac{8^x+2^x+1}{1-3*2^{3x}}}\\~\\


=\lim\limits_{x\to\infty}~\sqrt[3]{\dfrac{~~\dfrac{8^x+2^x+1}{8^x}~~}{\dfrac{1-3*2^{3x}}{8^x}}}\\~\\

\]

Answer 8

I did that but still wasn't able to evaluate it. I understood what you're saying but when I implemented it, it's not really working right for me.

Answer 9

\[\lim\limits_{x\to\infty}~\sqrt[3]{\dfrac{8^x+2^x+1}{1-3*2^{3x}}}\\~\\


=\lim\limits_{x\to\infty}~\sqrt[3]{\dfrac{~~\dfrac{8^x+2^x+1}{8^x}~~}{\dfrac{1-3*2^{3x}}{8^x}}}\\~\\

=\lim\limits_{x\to\infty}~ \sqrt[3]{\dfrac{1+\dfrac{1}{4^x} +\dfrac{1}{8^x} }{ \dfrac{1}{8^x} -3 }}\\~\\

= \sqrt[3]{\lim\limits_{x\to\infty}~\dfrac{1+\dfrac{1}{4^x} +\dfrac{1}{8^x} }{ \dfrac{1}{8^x} -3 }}\\~\\

\]

Answer 10

take the limit

Answer 11

\[ = \sqrt[3]{\dfrac{1+0+0}{0-3}}\]

Answer 12

simplify

Answer 13

thank u so much! I had gone to ask if perhaps the question was printed wrong. Turns out I was right. Thank u for your help and everyone else too. =)