Question

lim x^2 +2x-15 / x^2-x-6
x→3

Answer 1

Answer:

8/5

Step-by-step explanation:

3^2+2(3)-15/3^2-3-6

9+6-15/9-3-6

0/0

now

lim2x+2/2x-1

x-3

2(3)+2/2(3)-1

8/5

answer=8/5

Answer 2

Solution :

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6}}$

$\textsf{By factorizing the Numerator and Denominator, we get :}$

$\underline{\bf{Numerator}}\quad|\quad\underline{\bf{Denominator}} \\ \\ \sf{x^{2} + 2x - 15} \quad|\quad \sf{x^{2} - x - 6} \\ \\ \textsf{By using the middle-splitting factor theorem, we get :-} \\ \\ \sf{x^{2} + 2x - 15} \quad|\quad \sf{x^{2} - x - 6} \\ \\ \sf{x^{2} + (5 - 3)x - 15} \quad|\quad \sf{x^{2} - (3 - 2)x - 6} \\ \\ \sf{x^{2} + 5x - 3x - 15} \quad|\quad \sf{x^{2} - 3x + 2x - 6} \\ \\ \sf{x(x + 5) - 3(x + 5)} \quad|\quad \sf{x(x - 3) + 2(x - 3)} \\ \\ \sf{(x - 3)(x + 5)} \quad|\quad \sf{(x - 3)(x + 2)} \\ \\ \textsf{Hence, the numerator is (x - 3)(x + 5)} \\ \textsf{And the denominator is (x - 3)(x + 2)} \\ \\ \textsf{So, by substituting them in the equation, we get :-}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \lim_{x \to 3} \dfrac{\cancel{(x - 3)}(x + 5)}{\cancel{(x - 3)}(x + 2)}}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \lim_{x \to 3} \dfrac{(x + 5)}{(x + 2)}}$

$\textsf{Now, by using the quotient rule of limits, we get :} \\ \\ \underline{\sf{\bigstar\: Quotient\:rule\:of\:limits :-}} \\ \\ :\implies \sf{\lim_{x \to a} \dfrac{f(x)}{g(x)} = \dfrac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)}}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \dfrac{\lim_{x \to 3} (x + 5)}{\lim_{x \to 3} (x + 2)}}$

$\textsf{By using the sum rule of limits, we get :} \\ \\ \underline{\sf{\bigstar\:Sum\:rule\:of\:limits :-}} \\ \\ :\implies \sf{\lim_{x \to a} [f(x) + g(x)] = \lim_{x \to a} f(x) + \lim_{x \to a} g(x)}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \dfrac{\lim_{x \to 3} x + \lim_{x \to 3} 5}{\lim_{x \to 3} x + \lim_{x \to 3} 2}}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \dfrac{3 + 5}{3 + 2}}$

$:\implies \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \dfrac{8}{5}}$

$\therefore \sf{\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6} = \dfrac{8}{5}}$

$\sf{Hence,\:the\:value\:of\:\lim_{x \to 3} \dfrac{x^{2} + 2x - 15}{x^{2} - x - 6}\:is\:\dfrac{8}{5}}$