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

  1. $1$
  2. $53/12$
  3. $108/7$
  4. Limit does not exist
asked in Calculus by Veteran (395k points)
edited by | 1.5k views

5 Answers

+2 votes
Best answer

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

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

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

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

Put the limit and get the value

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

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

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

answered by Boss (34.3k points)
selected by
+6 votes

Use  L'Hospital's Rule  => 108/7

answered by Loyal (7.1k points)
+2 votes
y = $ \lim_{ x \rightarrow 3}  \left (  \frac{ x^{4} - 81 } { 2 x^{2} -5x -3 } \right )$

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

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

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

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

  = $\left ( \frac{6*18}{7} \right )$

  =$ \left ( \frac{108 } { 7 } \right )$
answered by Loyal (5.9k points)
0
While factorizing the denominator suppose i write it as (x-3)(x+1/2)
Now the answer comes as 216/7
0

(x-3)(x+1/2)

@okkpiyush

solving it is giving $\frac{1}{2} (2x^{2}-5x -3)$

I think u missed 1/2.  (216/2 = 108)

silly mistake.

+1 vote
108/7 so, ans is c
answered by Junior (525 points)
0 votes

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 $\frac{108}{7}$

answered by Junior (619 points)
Answer:

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