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

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1. anonymous

the limit =-3

2. anonymous

$\lim_{x \rightarrow 0}\frac{ 2x -3cosx}{ 1}$

3. anonymous

@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. anonymous

these problem can be worked out just by looking at them

5. anonymous

but it seems even when you tell the rules they repeat same question over and over again

6. anonymous

Just keep explaining will help. Sooner or later, the understanding will arrive...