Question

evaluate the limit of the following question.​

Answer 1

Answer:

(x-4) (x²-1)/x²-3x-4

(x-4){(x)²-(1)²}/x²-(4-1)x-4

(x-4)(x+1)(x-1)/x²-4x+x-4

(x-4)(x+1)(x-1)/x(x-4)+1(x-4)

(x-4)(x+1)(x-1)/(x-4)(x+1)

(x-1)

Step-by-step explanation:

hope it's helps you

Answer 2

☯ Explanation,

$\tt \red \bigstar \: lim_{x \rightarrow \: 4} \bigg( \dfrac{ (x - 4)( x{}^{2} - 1) }{x {}^{2} - 3x - 4 } \bigg) \\ \\ \\ \tt \underline{Applying \: L - Hospital \: rule, } \\ \\ \\ \leadsto \tt \: lim_{x \rightarrow \: 4} \bigg( \dfrac{ \dfrac{d}{dx}(x - 4)(x {}^{2} - 1) }{ \dfrac{d}{dx}(x {}^{2} - 3x - 4)} \bigg) \\ \\ \\ \leadsto \tt \: lim_{x \rightarrow \: 4} \bigg( \dfrac{3x {}^{2} - 8x - 1 }{2x - 3} \bigg) \\ \\ \\ \leadsto \tt \: \bigg( \dfrac{3 \times 4 {}^{2} - 8 \times 4 - 1 }{2 \times 4 - 3} \bigg) \\ \\ \\ \leadsto \tt \: \bigg( \dfrac{48 - 32 - 1}{8 - 3} \bigg) \\ \\ \\ \leadsto \tt \: \bigg( \not\dfrac{15}{5} \bigg) \\ \\ \\ \leadsto { \underline{\tt{ \boxed{3}}}} \: \green \bigstar$

☯ Hence,

$\dag \boxed{ \tt{lim_{x \rightarrow \: 4} \bigg( \dfrac{ (x - 4)( x{}^{2} - 1) }{x {}^{2} - 3x - 4 } \bigg) = 3}}$