Answer
$\lim\limits_{x \to 1^+}\frac{x^2-5x+6}{x-1}=\infty \\
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Work Step by Step
$\lim\limits_{x \to 1^+}\frac{x^2-5x+6}{x-1}=\infty \\
(as\,x\,approach\,1\,from\,right\,\,\,the\,numerator\,\\(x^2-5x+6)\,approach\,\,2\,\\and\,\,\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
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