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How do you show the limit does not exist limx→4x−4x2−8x+16lim_(x->4)(x-4)/(x^2-8x+16) ?

User Bobbin
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1 Answer

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\lim\limits_(x\to4)(x-4)/(x^2-8x+16)=\lim\limits_(x\to4)(x-4)/((x-4)^2)=\lim\limits_(x\to4)(1)/(x-4)\\\\\lim\limits_(x\to4^-)(1)/(x-4)=(1)/(0^-)=-\infty\\\\\lim\limits_(x\to4^+)(1)/(x-4)=(1)/(0^+)=\infty
User Mahmoud Ali
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