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

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 $\frac{108}{7}$

Answer 2

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.

Answer 3

108/7