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The problem is: $\lim_{x \to 0} \frac{1-4^x}{3^x-6^x}$ and I know from successive approximation and graphing that the answer is $2$. The student I am helping is in the beginning of Calculus so I cannot use anything but elementary limit laws. Thanks for your thoughts.

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  • $\begingroup$ I think I typed it wrong (1-4^x)/((3^x)-6^x)) $\endgroup$ Commented Feb 22, 2016 at 16:52

2 Answers 2

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Let us factor the numerator and denominator of the expression so we can eliminate the problem.

We have $$\lim_{x \to 0} \frac{1 - 4^{x}}{3^{x} - 6^{x}}$$ $$= \lim_{x \to 0} \frac{\left(1 + 2^{x}\right)\left(1 - 2^{x}\right)}{3^{x}\left(1 - 2^{x}\right)}$$ $$= \lim_{x \to 0} \frac{1 + 2^{x}}{3^{x}}$$ $$= \boxed{2}.$$

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Call the value of the limit $\ell$. Write the Taylor expansion of each term around $0$ as $4^x = 1 + x\log 4 + \cdots$ and etc. to get $$\ell = \lim_{x \to 0} \frac {- \log 4 (x + x^2/2 + \cdots)} {(\log 3 - \log 6) (x + x^2/2 + \cdots)} = \frac {\log 4} {\log 6 - \log 3} = \frac {2 \log 2} {\log 2} = 2.$$

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