Evaluate limx→−∞ln(−x3+x).
I was wondering if I can solve this limit in this way:
limx→−∞ln(−x3+x)=limx→−∞ln[x3(1+1x2)].
At this point, I just considered ln(x3) because 1 doesn't make any difference and 1/x2 tends towards 0.
So, the result will be 0. And I found it because I know the graph of the logarithm of x to the power of an odd number. So, my second question is, is it possible to understand what the result of limx→−∞ln(x3) algebraically without thinking of the graph?
Any suggestion and help will be appreciated.
Thank you in advance and have a good day :)
Answer
The correct way is
limx→−∞ln(−x3+x)=limx→−∞ln[−x3(1−1x2)]=limx→−∞[ln−x3+ln(1−1x2)]==limx→−∞ln−x3+limx→−∞ln(1−1x2)=+∞+0=+∞
indeed
limx→−∞ln−x3=limx→−∞3ln−x=limy→+∞3lny=+∞
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