Answer to Question #186262 in Calculus for noman

Question #186262

Is the function h(u,v)=cos⁡(u)+e^nu+√(sinh⁡(u^2 )+√(v+e^nu ))   , Where n is second digit of your arid number for example if your arid number is 19-arid-12345 then choose n=2. Is is continuous? In case of yes then find the points of continuity. Explain briefly


1
Expert's answer
2021-05-07T09:30:37-0400

"\\displaystyle\nh(u,v)=cos\u2061(u)+e^{2u}+\\sqrt{\\left(\\sinh\u2061(u^2 )+\\sqrt{v+e^{2u}}\\right)}\\\\\n\n\\textsf{For}\\,\\, h(u, v)\\,\\, \\textsf{to continuous,}\\,\\, h(u, v)\\\\\n\\textsf{has to be defined}\\\\\n\\textsf{and for the function to be defined,}\\\\\n\nv + e^{2u} > 0\\,\\, \\\\\n\\textsf{to make the the function}\\\\\\textsf{in the square root positive}\\\\\n\n\\implies v > -e^{2u}\\\\\n\n\n\\textsf{Also, if}\\,\\, z = \\sqrt{v + e^{2u}}\\\\\n\n \\sinh(u^2) + z > 0 \\\\\n\n\\sinh(u^2) > -z\\\\\n\nu^2 > \\arcsin\\mathrm{h}\\left(-\\sqrt{v + e^{2u}}\\right)\\\\\n\n\\textsf{But}\\,\\, v + e^{2u} > 0,\\\\\n\n\\implies u^2 > \\arcsin\\mathrm{h}\\left(\\sqrt{v + e^{2u}}\\right) > \\arcsin\\mathrm{h}(0) = 0\\\\\n\n\\implies u > 0\\\\\n\n\\therefore h(u, v) \\,\\, \\textsf{is continuous if and only if}\\\\ \nu >0\\,\\, \\textsf{and}\\,\\, v > -e^{2u}."


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