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

ZeHanzBest 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.
 one year ago

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

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

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