Limit

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[math]\lim_{x\to1}(x^2-5x+5)^{\frac{2x+4}{x^3-x}}=\lim_{x\to1}e^{\frac{2x+4}{x^3-x}\ln(x^2-5x+5)}\\\\\text{Is this a right track? What should I do next?}[/math]

 

[BigBeachBanana]

You’re on the right track. In the exponent, you have indeterminate form of 0/0. Use l’hospital.

 

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[math]\lim_{x\to1}(x^2-5x+5)^{\frac{2x+4}{x^3-x}}=\\\\\lim_{x\to1}e^{\frac{(2x+4)\ln(x^2-5x+5)}{x^3-x}}=\\\\\lim_{x\to1}e^{\frac{2\ln(x^2-5x+5)+\frac{(2x+4)(2x-5)}{x^2-5x+5}}{3x^2-1}}=\\\\=e^\frac{0+6\cdot(-3)}{3-1}=e^{-9}\\\\\text{Thank you sir, that's my answer :)}[/math]