The discussion focuses on finding the limit of the expression $$\large e^{\lim_{x\to0}\frac1x\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}$$. A participant suggests rewriting the limit as $$\lim_{x\to0}\frac{\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}{x}$$ and applying l'Hôpital's Rule to evaluate it. The final step involves taking the exponential of the limit result to obtain the answer.
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Having got as far as $$\large e^{\lim_{x\to0}\frac1x\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}$$, you could write the limit as $$\lim_{x\to0}\frac{\log\left(\frac{1^{x+1} + 2^{x+1} + 4^{x+1}}7\right)}x$$ and use l'Hopital. (When you have found that limit, don't forget to take the exponential in order to get the answer.)goody said: