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For example:

$$\lim_{x\to \infty}\frac{2\cdot7^x -3\cdot5^x}{2\cdot3^x+4\cdot7^x}$$

I know that both the numerator and the denominator diverge to infinity, so my first instinct is to use L'Hopital's rule, and take the derivative of the numerator and denominator separately. But this clearly doesn't accomplish much for an exponential:

$$\lim_{x\to \infty}\frac{\ln(49)\cdot7^x-\ln(125)\cdot5^x}{\ln(9)\cdot3^x+\ln(7^4)\cdot7^x}$$

Plugging into Wolfram Alpha, the answer is apparently $\frac{1}{2}$. What is the process for solving this limit?

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1 Answer 1

up vote 4 down vote accepted

$$\lim_{x\to \infty}\frac{2\cdot7^x -3\cdot5^x}{2\cdot3^x+4\cdot7^x}$$

$$=\lim_{x\to \infty}\frac{2 -3\cdot\left(\frac57\right)^x}{2\cdot\left(\frac37\right)^x+4}=\frac24$$

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Thank you for your prompt answer! –  Ataraxia Dec 28 '12 at 20:43
    
@phoenixheart6, my peasure. –  lab bhattacharjee Dec 29 '12 at 5:08

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