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

ZeHanz
 2 years 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
 2 years ago
Best ResponseYou've already chosen the best response.0these problem can be worked out just by looking at them

mathsmind
 2 years 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
 2 years 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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