limit as x tends to 2,(root of 6-x, minus 2) all over (root of 3-x, minus 1)
"lim_{x\\to 2}\\frac{\\sqrt{6-x}-2}{\\sqrt{3-x}-1}\\newline\nPutting x=2,\\newline\n=\\frac{\\sqrt{4}-2}{\\sqrt{1}-1}\\newline\n=\\infty"
Thus, the limit of the given function is "\\infty" .
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