Question

pls answer
plssss. as soon as possible​

Answer 1

$\huge{\pink{\boxed{\boxed{\boxed{\purple{\underline{\underline{\green{\mathbf{THANKS}}}}}}}}}}$

Answer 2

On direct checking,

$\quad$

$\displaystyle\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\dfrac {(3)^4-81}{2(3)^2-5(3)-3}\\\\\\\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\dfrac {0}{0}$

$\quad$

Then,

$\quad$

$\displaystyle\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\lim_{x\to3}\dfrac {x^4-3^4}{2x^2+x-6x-3}\\\\\\\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\lim_{x\to3}\dfrac {(x-3)(x+3)(x^2+3^2)}{(x-3)(2x+1)}\\\\\\\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\lim_{x\to3}\dfrac {(x+3)(x^2+3^2)}{2x+1}\\\\\\\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\dfrac {(3+3)(3^2+3^2)}{2\cdot3+1}\\\\\\\underline {\underline {\lim_{x\to3}\dfrac {x^4-81}{2x^2-5x-3}=\dfrac {108}{7}}}$