Here's the question you clicked on:
iloveyahhhh
Find the limit as x approaches 0 for (x^2-3sinx)/x
\[\lim_{x \rightarrow 0}\frac{ 2x -3cosx}{ 1}\]
@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 well-known standard limit (with an outcome of 1) So the result would be 0-3*1=-3.
these problem can be worked out just by looking at them
but it seems even when you tell the rules they repeat same question over and over again
Just keep explaining will help. Sooner or later, the understanding will arrive...