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Compute $\displaystyle \lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3}$

  1. $1$
  2. $53/12$
  3. $108/7$
  4. Limit does not exist
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7 Answers

Best answer
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Let $y=\displaystyle\lim_{x\rightarrow3}\frac{x^{4}-81}{2x^{2}-5x-3}$

When we put $3$ in the equation we get $\frac{0}{0}$ form, so we can apply L Hospital$'$s rule.

Differentiate the numerator and denominator separately

$y=\displaystyle\lim_{x\rightarrow3}\frac{4x^{3}-0}{4x-5-0}$

$y=\displaystyle\lim_{x\rightarrow3}\frac{4x^{3}}{4x-5}$

Put the limit and get the value

$y=\frac{4\times(3)^{3}}{4\times(3)-5}$

$y=\frac{4\times27}{7}$

$y=\frac{108}{7}$

Correct Answer is C.

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y = $ \lim_{ x \rightarrow 3}  \left (  \frac{ x^{4} - 81 } { 2 x^{2} -5x -3 } \right )$

  = $\lim_{ x \rightarrow 3} \left ( \frac{(x^{2})^{2}-9^{2}}{2x^{2}-5x-3}\right )$

  = $\lim_{ x \rightarrow 3} \left (\frac{(x-3)(x+3)(x^{2}+9)}{(2x+1)(x-3)}\right )$

  = $\lim_{ x \rightarrow 3} \left (\frac{(x+3)(x^{2}+9)}{(2x+1)}\right )$ (when $x\neq 3$)

  = $\left ( \frac{(3+3)(3^{2}+9)}{(2*3 +1)} \right )$

  = $\left ( \frac{6*18}{7} \right )$

  =$ \left ( \frac{108 } { 7 } \right )$
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