Limit, x=>+infinity, ln(3^x - x) [ ln(x^4 + 1) - 2 ln(x^2 - x) ]

Help for:

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \ln\left(3^x\, -\, x\right)\, \big[\ln\left(x^4\, +\, 1\right)\, -\, 2\, \ln\left(x^2\, -\, x\right)\big]\)

Noismaker said:

Help for:

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \ln\left(3^x\, -\, x\right)\, \big[\ln\left(x^4\, +\, 1\right)\, -\, 2\, \ln\left(x^2\, -\, x\right)\big]\)

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Have you considered rules of Logarithms? Show us where that gets you.

Noismaker said:

Help for:

. . . . .\(\displaystyle \displaystyle \lim_{x \rightarrow \infty}\, \ln\left(3^x\, -\, x\right)\, \big[\ln\left(x^4\, +\, 1\right)\, -\, 2\, \ln\left(x^2\, -\, x\right)\big]\)

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What have you covered recently in class? (That is, what tools are you probably expected to use for this exercise?) I mean, yes, we can find the limit by other means, but you can't likely expect your instructor to "trust you for it"; ya gotta show yer work!

When you reply, please include a clear listing of your thoughts and efforts so far. Thank you!