limit as x approaches 0 of cot(pi/2-x)

{ "query": { "display": "$$\\lim_{x\\to\\:0}\\left(\\cot\\left(\\frac{π}{2}-x\\right)\\right)$$", "symbolab_question": "BIG_OPERATOR#\\lim _{x\\to 0}(\\cot(\\frac{π}{2}-x))" }, "solution": { "level": "PERFORMED", "subject": "Calculus", "topic": "Limits", "subTopic": "SingleVar", "default": "0", "meta": { "showVerify": true } }, "steps": { "type": "interim", "title": "$$\\lim_{x\\to\\:0}\\left(\\cot\\left(\\frac{π}{2}-x\\right)\\right)=0$$", "input": "\\lim_{x\\to\\:0}\\left(\\cot\\left(\\frac{π}{2}-x\\right)\\right)", "steps": [ { "type": "step", "primary": "Plug in the value $$x=0$$", "result": "=\\cot\\left(\\frac{π}{2}-0\\right)", "meta": { "title": { "extension": "Limit properties - if the limit of f(x), and g(x) exists, then:<br/>$$\\bullet\\quad\\lim_{x\\to\\:a}\\left(x\\right)=a$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[c\\cdot{f\\left(x\\right)}]=c\\cdot\\lim_{x\\to{a}}{f\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[\\left(f\\left(x\\right)\\right)^c]=\\left(\\lim_{x\\to{a}}{f\\left(x\\right)}\\right)^c$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\pm{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\pm\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}[f\\left(x\\right)\\cdot{g\\left(x\\right)}]=\\lim_{x\\to{a}}{f\\left(x\\right)}\\cdot\\lim_{x\\to{a}}{g\\left(x\\right)}$$<br/>$$\\bullet\\quad\\lim_{x\\to{a}}\\left(\\frac{f\\left(x\\right)}{g\\left(x\\right)}\\right)=\\frac{\\lim_{x\\to{a}}{f\\left(x\\right)}}{\\lim_{x\\to{a}}{g\\left(x\\right)}},\\:$$where $$\\lim_{x\\to{a}}g\\left(x\\right)\\neq0$$" } } }, { "type": "interim", "title": "Simplify $$\\cot\\left(\\frac{π}{2}-0\\right):{\\quad}0$$", "input": "\\cot\\left(\\frac{π}{2}-0\\right)", "result": "=0", "steps": [ { "type": "step", "primary": "$$\\frac{π}{2}-0=\\frac{π}{2}$$", "result": "=\\cot\\left(\\frac{π}{2}\\right)" }, { "type": "step", "primary": "Use the following trivial identity:$${\\quad}\\cot\\left(\\frac{π}{2}\\right)=0$$", "secondary": [ "$$\\cot\\left(x\\right)$$ periodicity table with $$πn$$ cycle:<br/>$$\\begin{array}{|c|c|}\\hline x&\\cot(x)\\\\\\hline 0&\\mp\\infty\\\\\\hline \\frac{π}{6}&\\sqrt{3}\\\\\\hline \\frac{π}{4}&1\\\\\\hline \\frac{π}{3}&\\frac{\\sqrt{3}}{3}\\\\\\hline \\frac{π}{2}&0\\\\\\hline \\frac{2π}{3}&-\\frac{\\sqrt{3}}{3}\\\\\\hline \\frac{3π}{4}&-1\\\\\\hline \\frac{5π}{6}&-\\sqrt{3}\\\\\\hline \\end{array}$$" ], "result": "=0" } ], "meta": { "solvingClass": "Solver", "interimType": "Generic Simplify Specific 1Eq", "gptData": "pG0PljGlka7rWtIVHz2xymbOTBTIQkBEGSNjyYYsjjDErT97kX84sZPuiUzCW6s7hxDC+iBQjKnYgmb+i2+ULSq5ZgL0nIFjp3BI7L1e/jN1g99dC9fj9sg0EHzBIRDRd79UrkSVT0SCLs80Lgihlx429vuTSxWa7B/X3D1oP01kp49jvQW/qCTjuvhVJzjMZhG9q9lM95MzSFoL+F0aBb8yD3hLQ33B7/8/LpbPE3o=" } } ], "meta": { "solvingClass": "Limits", "practiceLink": "/practice/limits-practice", "practiceTopic": "Limits" } }, "plot_output": { "meta": { "plotInfo": { "variable": "x", "plotRequest": "yes" }, "showViewLarger": true } }, "meta": { "showVerify": true } }