Answer
$\lim\limits_{x \to 0^+}(-10\cot x)=\lim\limits_{x \to 0^+}\frac{-10\cos x}{\sin x}=-\infty \\
(as\,x\,approach\,0\,from\,right\,\,\,\cos x\,approach\,1\,so\,the\,numerator\,approach\,\,-10\,\\and\,\sin x\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
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Work Step by Step
$\lim\limits_{x \to 0^+}(-10\cot x)=\lim\limits_{x \to 0^+}\frac{-10\cos x}{\sin x}=-\infty \\
(as\,x\,approach\,0\,from\,right\,\,\,\cos x\,approach\,1\,so\,the\,numerator\,approach\,\,-10\,\\and\,\sin x\,is\,positive\,and\,approach\,0 \,\\so\,the\,denominator\,is\,positive\,and\,approach\,\,\,0)\\
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