I am trying to solve this question but I am getting a negative infinity which is wrong. $$\lim _{x\to \infty }\left(\ln(e^{2x}-1)-\frac{x^2-3}{x}\right)$$ $$=\lim _{x\to \infty }\left(\ln(e^{2x}-1)-\frac{x(x^{ }-\frac{3}{x})}{x}\right)$$ $$=\lim _{x\to \infty }\left(\ln(e^{2x}-1)-x\right)$$ $$=\lim \:_{x\to \:\infty \:}\left((e^{2x}-1)\frac{\ln(e^{2x}-1)}{e^{2x}-1}-x\right)$$ $$\lim \:_{x\to \:\infty \:}\:(-x)$$ $$=-\infty $$
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1How did you get from Step 2 to Step 3? Also, you have too many brackets for each step to make sense – Adam Rubinson Feb 03 '20 at 17:36
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And how do you get from step 4 to step 5? I don't think your cancellations make sense. – xxxxxxxxx Feb 03 '20 at 17:37
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sir actually 3/x will go zero as i am neglecting and also the remaining x(x)/x will cancel with giving only result x – Feemo Fellow Feb 04 '20 at 00:12
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Just split up the $e^{2x}$ and you get
$$\ln(e^{2x}-1) - x+\frac 3x \stackrel{e^{2x}-1=e^{2x}(1-e^{-2x})}{=} \ln(e^{2x}) + \ln(1-e^{-2x}) - x+\frac 3x$$ $$=2x-x+\frac 3x + \ln(1-e^{-2x})=x+\frac 3x + \ln(1-e^{-2x})\stackrel{x \to +\infty}{\rightarrow}+\infty$$
trancelocation
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actually answer log2 but when i put into desmos this function it actually does not exits – Feemo Fellow Feb 04 '20 at 00:14
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this question even my sir ask me to put on his email still i am waiting for his reply – Feemo Fellow Feb 04 '20 at 00:16
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@FeemoFellow : Maybe the graph can convince you that the limit is $+\infty$: https://www.wolframalpha.com/input/?i=plot+%28%5Cln%28e%5E%7B2x%7D-1%29-%5Cfrac%7Bx%5E2-3%7D%7Bx%7D%29%2C+x%3D0..20 – trancelocation Feb 04 '20 at 04:07