Question

Evaluate the limit (x-4)/(x2 - x - 12) as x approaches 4.
• A. O
B. undefined
C. 1/7
D. infinity
can I get the explanation​

Answer 1

$\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} - x - 12}$

If we put x = 4 in (x-4)/(x2 - x - 12) then we will get 0/0 form. So we will further simplify the given expression.

$\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} -4 x + 3x - 12}$

[ -4x × 3x = -12x² , 3x-4x = -x ]

$\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{x( {x} -4 ) + 3(x - 4)}$

$\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{( {x} + 3 )(x - 4)}$

$\implies\displaystyle \sf \lim \limits_{ {x \to \: 4}} \: \dfrac{ \cancel{(x - 4)}}{( {x} + 3 ) \: \: \cancel{(x - 4)}}$

$\implies\displaystyle \sf \lim \limits_{ {x \to \: 4}} \: \dfrac{ {1}}{( {x} + 3 ) {}}$

Now put x =4 to get final answer.

$\implies\displaystyle \sf \dfrac{ {1}}{ {4} + 3}$

$\implies\displaystyle \sf \dfrac{ {1}}{ 7}$

Answer :

$\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} - x - 12} = \dfrac{1}{7}$

Therefore option (C) is correct.

Answer 2

Answer:

HELLO HERE IS YOUR ANSWER

\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} - x - 12}x→4limx2−x−12x−4

If we put x = 4 in (x-4)/(x2 - x - 12) then we will get 0/0 form. So we will further simplify the given expression.

\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} -4 x + 3x - 12}⟹x→4limx2−4x+3x−12x−4

[ -4x × 3x = -12x² , 3x-4x = -x ]

\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{x( {x} -4 ) + 3(x - 4)}⟹x→4limx(x−4)+3(x−4)x−4

\implies\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{( {x} + 3 )(x - 4)}⟹x→4lim(x+3)(x−4)x−4

\implies\displaystyle \sf \lim \limits_{ {x \to \: 4}} \: \dfrac{ \cancel{(x - 4)}}{( {x} + 3 ) \: \: \cancel{(x - 4)}}⟹x→4lim(x+3)(x−4)(x−4)

\implies\displaystyle \sf \lim \limits_{ {x \to \: 4}} \: \dfrac{ {1}}{( {x} + 3 ) {}}⟹x→4lim(x+3)1

Now put x =4 to get final answer.

\implies\displaystyle \sf \dfrac{ {1}}{ {4} + 3}⟹4+31

\implies\displaystyle \sf \dfrac{ {1}}{ 7}⟹71

Answer :

\displaystyle \sf \lim \limits_{x \to \: 4} \: \dfrac{ x - 4}{ {x}^{2} - x - 12} = \dfrac{1}{7}x→4limx2−x−12x−4=71

Therefore option (C) is correct.