what is the limit as x approaches to 1+ of (x^3-1)ln(x-1)^5
1 Answer
We have
\[\lim_{x\rightarrow 1^+} (x^3-1)\ln (x-1)^5=\lim_{x\rightarrow 1^+} \frac{\ln (x-1)^5}{\frac{1}{x^3-1}}\]
\[=\lim_{x\rightarrow 1^+} \frac{5\ln (x-1)}{\frac{1}{x^3-1}}=\lim_{x\rightarrow 1^+} \frac{5\frac{1}{x-1}}{\frac{-3x^2}{(x^3-1)^2}}\]
\[=\lim_{x\rightarrow 1^+} \frac{5(x^3-1)^2}{-3x^2(x-1)}\]
\[=\lim_{x\rightarrow 1^+} \frac{5 [(x-1)(x^2+x+1)]^2}{-3x^2(x-1)}=\lim_{x\rightarrow 1^+} \frac{5 (x-1)^2(x^2+x+1)^2}{-3x^2(x-1)}\]
\[=\lim_{x\rightarrow 1^+} \frac{5 (x-1)(x^2+x+1)^2}{-3x^2}=0.\]
Daniel90
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