Compute $\displaystyle \lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}$
Answer is: C
On Putting the value of limit in Numerator and Denominator it comes out to be $\frac{0}{0}$
Applying L'Hopital Rule $\[ \lim_{x \to 3} f(x) /g(x) \]= \[ \lim_{x \to 3} f'(x)/g'(x)) \]\ =$ $\frac{4x^{3}}{4x-5}$
On putting value x=3 $\frac{108}{7}$
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