Question

⇒ Evaluate the following limit they exist.​

Answer 1

Kindly follow the attachment for answer!

Answer 2

ANSWER:

To Evaluate:

$\:\:\:\:\bullet\:\:\:\:\displaystyle\lim_{x\to3}\:\dfrac{x^3-4x-15}{x^3+x^2-6x-18}$

Solution:

$\displaystyle:\longrightarrow\lim_{x\to3}\:\dfrac{x^3-4x-15}{x^3+x^2-6x-18}\\\\\text{Putting x = 3}\\\\:\implies\dfrac{(3)^3-4(3)-15}{(3)^3+(3)^2-6(3)-18}\\\\:\implies\dfrac{27-12-15}{27+9-18-18}\\\\:\implies\dfrac{27-27}{36-36}\\\\:\implies\dfrac{0}{0}\\\\\text{As, 0/0 is undefined, we apply L'Hospital's Rule}$

L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$\text{So,}\\\\:\implies\displaystyle\lim_{x\to3}\:\dfrac{x^3-4x-15}{x^3+x^2-6x-18}\\\\:\implies\displaystyle\:\lim_{x\to3}\dfrac{\dfrac{d}{dx}x^3-4x-15}{\dfrac{d}{dx}x^3+x^2-6x-18}\\\\\text{We know that,}\\\\:\implies\dfrac{d}{dx}x^n=nx^{n-1}\:\:\:\:\&\:\:\:\:\dfrac{d}{dx}constant=0\\\\\text{So,}\\\\:\implies\displaystyle\lim_{x\to3}\:\dfrac{\dfrac{d}{dx}x^3-4x-15}{\dfrac{d}{dx}x^3+x^2-6x-18}\\\\:\implies\displaystyle\lim_{x\to3}\:\dfrac{3x^2-4}{3x^2+2x-6}\\\\\text{Now, putting x = 3}\\\\:\implies\dfrac{3(3)^2-4}{3(3)^2+2(3)-6}\\\\:\implies\dfrac{3(9)-4}{3(9)+6-6}\\\\:\implies\dfrac{27-4}{27-0}\\\\\bf{:\implies\dfrac{23}{27}}$

Formulae Used:

• L'Hospital's Rule : It states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
• d/dx x^n = nx^{n-1}
• d/dx constant = 0