MHB Finding Limit: Get Step-by-Step Help Here

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To find the limit, start with the expression e^(lim(x→0)(1/x)log((1^(x+1) + 2^(x+1) + 4^(x+1))/7)). Rewriting the limit as lim(x→0)(log((1^(x+1) + 2^(x+1) + 4^(x+1))/7)/x) allows the application of l'Hôpital's rule. After calculating the limit, remember to exponentiate the result to obtain the final answer. This method effectively addresses the divergence issue encountered with 1/x. Following these steps will lead to the correct limit evaluation.
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Hello everyone, I need to find this limit
limitt.png
. What I tried is that
limit.png
,
but clearly, 1/x diverges so I don't think it was very helpful.
Could someone help me what I need to do please?
 
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goody said:
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?
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.)
 

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