I need some help with a Limit

Vincent_M05

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Ok so I broke down the limit into 2 parts, in the U part, and in the V part I was really confident when I calculated the Limit for U and V and I thought that the +1 that was left would not bother me since it would just add 1 to the limit so 6-->7 but when I looked up the answer it was e^6. I am really stuck right now and I don't know why this would be the right answer, it 1:50 AM right now, so this could explain why I am so stuck on this question or I just have no clue what is going on since I have started learning about limits 3 days ago. I will attach my notes to this Thread and I would appreciate any help. Thank you!
 

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The [imath]2x[/imath] is in the exponent, not multiplication.
Rewrite the limit using exponential of logarithmic function and power rule:
[math]\lim_{x \to \infty}\left(1+\frac{3x}{x^2+2}\right)^{2x}= \lim_{x \to \infty}e^{2x\ln\left(1+\frac{3x}{x^2+2}\right)}[/math]
 
The [imath]2x[/imath] is in the exponent, not multiplication.
Rewrite the limit using exponential of logarithmic function and power rule:
[math]\lim_{x \to \infty}\left(1+\frac{3x}{x^2+2}\right)^{2x}= \lim_{x \to \infty}e^{2x\ln\left(1+\frac{3x}{x^2+2}\right)}[/math]

Thank you very much?. Could you give me a link for a website that explains about that topic (exponential of logarithmic function and power rule), that would be nice.
 
Thank you very much?. Could you give me a link for a website that explains about that topic (exponential of logarithmic function and power rule), that would be nice.
Since you're doing limits, I assume that you have been taught logarithms and exponents before. But if you need a refresher, here are some links:
 
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