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I'm having trouble finding the answer to the limit $$\lim\limits_{i\to\infty} \dfrac{5^i - 2^{3i+1}}{7^{i-2}+8^i}$$

I get the answer 98, which to me seem to be wrong. Can someone help me?

Thanks in advance.

  • 0
    Is it not 2? It looks like 2 – 2017-01-21
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    @Dr.MV wouldn't be -2? – 2017-01-21
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    @Ironping, sorry my bad. i was harping on 2(2^3i)/8^i – 2017-01-21
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    @SakethMalyala no problem thanks a lot for the help – 2017-01-21
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    @Ironping Yes, it is $-2$. – 2017-01-21

4 Answers 4

3

HINT:

Note first that $2^{3i+1}=2(8^i)$

Finish by dividing numerator and denominator by $8^i$ and letting $i \to \infty$.

2

In $a_i = \frac{5^i - 2^{3i+1}}{7^{i-2}+8^i} $, note that $2^{3i+1} =2^{3i}2 =2\cdot 8^i $.

Therefore $a_i = \frac{5^i - 2\cdot 8^i}{7^{i}/49+8^i} = \frac{(5/8)^i - 2}{(7/8)^{i}/49+1} \to -2 $ since $(5/8)^i \to 0$ and $(7/8)^i \to 0$.

2

Thanks everyone got it $a_i = \frac{5^i - 2^{3i+1}}{7^{i-2}+8^i}$ $$=\lim_{i\to\infty}\frac{5^i-2(8^i)}{7^i\cdot7^{-2}+8^i}\cdot\frac{\frac{1}{8^i}}{\frac{1}{8^i}}=\lim\limits_{i\to\infty} -2= -2 $$

0

Hint. One may write, as $i \to \infty$, $$ a_i = \frac{5^i - 2^{3i+1}}{7^{i-2}+8^i}=\frac{2^{3i}}{8^i}\cdot\frac{(5/8)^i - 2}{\frac1{49}(7/8)^{i}+1}. $$