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iloveyahhhh

  • 3 years ago

Find the limit as x approaches 0 for (x^2-3sinx)/x

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  1. mathsmind
    • 3 years ago
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    the limit =-3

  2. mathsmind
    • 3 years ago
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    \[\lim_{x \rightarrow 0}\frac{ 2x -3cosx}{ 1}\]

  3. ZeHanz
    • 3 years ago
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    @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.

  4. mathsmind
    • 3 years ago
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    these problem can be worked out just by looking at them

  5. mathsmind
    • 3 years ago
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    but it seems even when you tell the rules they repeat same question over and over again

  6. ZeHanz
    • 3 years ago
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    Just keep explaining will help. Sooner or later, the understanding will arrive...

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