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- Thread starter goody
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- Feb 7, 2012

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Having got as far as \(\displaystyle \large e^{\lim_{x\to0}\frac1x\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}\), you could write the limit as \(\displaystyle \lim_{x\to0}\frac{\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}x\) and use l'Hopital. (When you have found that limit, don't forget to take the exponential in order to get the answer.)Hello everyone, I need to find this limit View attachment 10284. What I tried is that

View attachment 10285,

but clearly, 1/x diverges so I don't think it was very helpful.

Could someone help me what I need to do please?