anonymous
  • anonymous
Find the limit as x approaches 0 for (x^2-3sinx)/x
Calculus1
katieb
  • katieb
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anonymous
  • anonymous
the limit =-3
anonymous
  • anonymous
\[\lim_{x \rightarrow 0}\frac{ 2x -3cosx}{ 1}\]
ZeHanz
  • ZeHanz
@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 well-known standard limit (with an outcome of 1) So the result would be 0-3*1=-3.

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anonymous
  • anonymous
these problem can be worked out just by looking at them
anonymous
  • anonymous
but it seems even when you tell the rules they repeat same question over and over again
ZeHanz
  • ZeHanz
Just keep explaining will help. Sooner or later, the understanding will arrive...

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