Limits question (w/natural logs)

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SUMMARY

The limit in question is limx→∞ (1+2x)11/(2ln(x)). As x approaches infinity, the term 2ln(x) also approaches infinity, leading to the conclusion that the limit approaches 0. However, the base (1+2x) grows significantly faster than the exponent (11/(2ln(x))), which complicates the evaluation. The correct approach involves rewriting the expression using the exponential function, specifically e11/(2ln(x))ln(1+2x), to analyze the limit accurately.

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Homework Statement


Find limit:
limx=> inf (1+2x)11/(2ln(x))

The Attempt at a Solution


Well, I have no real mathematical attempt to this. It was just my common sense that failed. Anyways, I figured that as x goes to infinity the 2ln(x) term will also go to infinity. Since that 11 on the top doesn't change, the term will eventually go to 0. So then we'd have (1+2*infinity) to the power of zero. Which should equal one.

HOWEVER, that's wrong. I think I know why - because the infinity in the base will be different from that in the power (the power one will be increasing slower because of the ln()). But I don't know. I know there should be some mathematical way of doing this, but I don't see. Any tips?
 
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ok i;ll give u a hint


[tex](1+2x)^\frac{11}{2ln(x)}=e^{\frac{11}{2ln(x)}ln(1+2x)}[/tex]
 

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