Question

plz help friends ...
solve both the problems .
please be specific and easy to understand .

Answer 1

According to L'Hospital's rule : If we have an indeterminate form (that is 0/0 or ∞/∞) then we need to differentiate the numerator and denominator and then apply the limit.

1. $\lim_{n \to 2} \frac{x^3 - 8}{x^2 - 4}$

here let us take numerator as f(x) = x³ - 8 and denominator as g(x) = x² - 4

⇒ f(2) = 2³ - 8 = 8 - 8 = 0

⇒ g(2) = 2² - 4 = 4 - 4 = 0

⇒ The given algebraic fraction is in the indeterminate form that is 0/0

Applying L'Hospital's rule :

Differentiating numerator that is f(x)

⇒ f(x) = x³ - 8

⇒ f¹(x) = 3x²

Differentiating denominator that is g(x)

⇒ g(x) = x² - 4

⇒ g¹(x) = 2x

Now Applying Limit we get :

$\lim_{n \to 2} \frac{3x^2}{2x}$

⇒ (3 × 2 × 2)/(2 × 2) = 3

2.  Numerator as f(x) = x³ - x² - 18 and denominator as g(x) = x - 3

⇒ f(3) = 3³ - 3² - 18 = 0

⇒ g(3) = 3 - 3 = 0

⇒ The given algebraic fraction is in the indeterminate form that is 0/0

Applying L'Hospital's rule :

Differentiating numerator that is f(x)

⇒ f(x) = x³ - x² - 18

⇒ f¹(x) = 3x² - 2x

Differentiating denominator that is g(x)

⇒ g(x) = x -3

⇒ g¹(x) = 1

Now applying limit :

$\lim_{n \to 2} 3x^2 - 2x$

⇒ 3(3 × 3) - 2(3)

⇒ 27 - 6 = 21