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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
in Calculus by Veteran (425k points)
edited by | 1.8k views

6 Answers

+8 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

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

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

$y=\frac{108}{7}$ Answer is (C)

by Veteran (54.9k points)
edited by
+6 votes

Use  L'Hospital's Rule  => 108/7

by Boss (15.2k points)
+4 votes
y = $ \lim_{ x \rightarrow 3}  \left (  \frac{ x^{4} - 81 } { 2 x^{2} -5x -3 } \right )$

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

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

  = $\lim_{ x \rightarrow 3} \left (\frac{(x+3)(x^{2}+9)}{(2x+1)}\right )$ (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 )$
by Boss (21.7k points)
edited by
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.

0

@Satbir you missed to write : $x \neq 3$ in the answer.

functions $\frac{(x-3)(x+3)(x^{2} + 9)}{(2x+1)(x-3)}$ and $\frac{(x+3)(x^{2} + 9)}{(2x+1)}$ are not same. Here, function $\frac{(x-3)(x+3)(x^{2} + 9)}{(2x+1)(x-3)}$ has a "hole" at $x=3$. These 2 functions will be same( wrt domain and codomain) when we don't consider the point $x=3$. It is like functions $\frac{x^2 - 9}{x-3}$ and $x+3$ are not same. Function  $\frac{x^2 - 9}{x-3}$ behaves exactly  as the straight line $y=x+3$ except  at the point $x=3$ where there exists a "hole".

0
That is the meaning of $lim _{x \rightarrow 3}$ right ?

we are considering cases when when x tends to 3 and NOT at x=3.

so both functions are same when $x!=3$
+1

so both functions are same when x!=3

yes, I agree, but this thing should be mentioned. You changed the original function to a new function without defined it.

It should be :

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

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

  This is a very small thing but most important and can be found in any book.

+1 vote
108/7 so, ans is c
by Junior (581 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 $\frac{108}{7}$

by Junior (825 points)
0 votes
first we will see condition of this function .

this is in the form of 0/0 so we will apply L-HOSPITAL rule

f(x)=f1(x)/f2(x)

derivative of f1(x)=4 x*3 = 4*3*3*3=108 at x=3

derivative of f2(x)=4x-5=7 at x=3.

f(x)=f1(3)/f2(3)= 108/27.
so option C is right.
by (87 points)

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