GATE CSE 2019 | Question: 13

Question

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

• A. $1$
• B. $53/12$
• C. $108/7$
• D. Limit does not exist

Comments

For those trying to "factorize" the denominator like I did, remember to make the leading coefficient $1$ first. You do this by writing the denominator as $2(x^2 - 5x/2 - 3/2)$.

Answer 1

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.

Comments

But to find its derivative it's in u/v form how can we derivative seperately? Please someone clarify

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Please read the concept: https://brilliant.org/wiki/lhopitals-rule/

Answer 2

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 )$

Comments

@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".

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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 $x!=3$

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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.

Answer 3

Use  L'Hospital's Rule  => 108/7

Answer 4

108/7 so, ans is c