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Homework Help
Calculus and Beyond Homework Help
Individually continuous function + monotonic = continious
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[QUOTE="esvee, post: 2745236, member: 213983"] [h2]Homework Statement [/h2] Given [tex]f:{R^2} \to R[/tex]. Prove that if [tex]f[/tex] is continuous individually for each variable, and monotone in the first variable, then [tex]f[/tex] is continuous. [h2]Homework Equations[/h2] [h2]The Attempt at a Solution[/h2] Well I "succeeded" to "prove" it by choosing [tex]\min \left( {{\varepsilon _1},{\varepsilon _2}} \right)[/tex] of the corresponding inequalities that follow from the definition of the limit. But I did not use the monotonicity at all, that means I got it wrong :) Explanation: If f is cont. relative to x: [tex] {\lim }\limits_{\left( {x,y} \right) \to \left( {{x_0},{y_0}} \right)} f\left( {x,y} \right) = f\left( {{x_0},y} \right) \Leftrightarrow \forall \varepsilon > 0,\exists {\delta _1} > 0,\left\| {\left( {x,y} \right) - \left( {{x_0},{y_0}} \right)} \right\| < {\delta _1} \Rightarrow \left| {f\left( {x,y} \right) - f\left( {{x_0},y} \right)} \right| < \varepsilon [/tex] etc. for y. I'm sure it must be some simple inequality trick or the like, please help me out people! [/QUOTE]
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Homework Help
Calculus and Beyond Homework Help
Individually continuous function + monotonic = continious
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