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mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0}\frac{ 2x 3cosx}{ 1}\]

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1@mathsmind has done it with l'Hôpital's Rule. That's ok. This one could also be done in a more basic way. Split up: \[\lim_{x \rightarrow 0}\frac{ x^2 }{ x }3\lim_{x \rightarrow 0}\frac{ \sin x }{ x }\] The left one is just\[\lim_{x \rightarrow 0}\frac{ x \cdot x }{ x }=\lim_{x \rightarrow 0}x=0\]The right one is 3 times the wellknown standard limit (with an outcome of 1) So the result would be 03*1=3.

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0these problem can be worked out just by looking at them

mathsmind
 one year ago
Best ResponseYou've already chosen the best response.0but it seems even when you tell the rules they repeat same question over and over again

ZeHanz
 one year ago
Best ResponseYou've already chosen the best response.1Just keep explaining will help. Sooner or later, the understanding will arrive...
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