Question 127

$$\lim_{x \rightarrow 3} \frac{x^3 - 3x^2}{3x^2 - 4x - 15} =$$

Solution

Solution:

As on putting limit ,$$\lim_{x \rightarrow 3}$$ f(x) -> $$\frac{0}{0}\ format$$

So using L hospitals rule

Differentiating Nr and Dr. $$\lim_{x \rightarrow 3}$$ $$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}$$

Putting the limits,

$$\frac{\left(3x^2\ -\ 6x\right)}{6x\ -\ 4}\ =\ \frac{\left(3.3^2\ -\ 6.3\right)}{6.3\ -\ 4}=\ \frac{\left(27-18\right)}{14}=\ \frac{9}{14}$$

$$\ \frac{9}{14}$$ Answer

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