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$$\lim_{x \rightarrow 3} \frac{x^3 - 3x^2}{3x^2 - 4x - 15} =$$
Solution:
As on putting limitΒ ,$$\lim_{x \rightarrow 3}$$ f(x) ->Β $$\frac{0}{0}\ format$$
So using L hospitals rule
Differentiating Nr and Dr.Β $$\lim_{x \rightarrow 3}$$Β Β $$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}$$
Putting the limits,Β
$$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}\ =\ \frac{\left(3.3^2\ -\ 6.3\right)}{6.3\ -\ 4}=\ \frac{\left(27-18\right)}{14}=\ \frac{9}{14}$$
$$\ \frac{9}{14}$$ Answer
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