Question

solve this by using factorisation method​

Answer 1

hey mate here is ur ans

Mark it as brainliest ans

Answer 2

$\Large{\underline{\underline{\mathfrak{\tt{\red{Solution}}}}}}$

Given here:-

$:\mapsto\tt{\orange{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{8x-3}{2x-1}-\dfrac{4x^2+1}{4x^2-1}\right)}} \\ \\ \\ \small\tt{\green{\:\:\:\:\:Factories\:these\:term}} \\ \\ \\ :\mapsto\tt{\:\displaystyle\lim_{x \to \frac{1}{2}} \left(\dfrac{8x-4}{2x-1}-\dfrac{4x^2+1}{(2x)^2-1^2)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{8x-4}{2x-1}-\dfrac{4x^2+1}{(2x+1)(2x-1)}\right)}$

$\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{(8x-3)(2x+1)-(4x^2+1)}{(2x-1)(2x+1)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{16x^2+8x-6x-3-4x^2-1}{(2x-1)(2x+1)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{12x^2+2x-4}{(2x+1)(2x-1)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{2(6x^2+x-2)}{(2x+1)(2x-1)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(\dfrac{2(6x^2+4x-3x-2)}{(2x+1)(2x-1)}\right)}$

$\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(2\times \dfrac{3x(2x-1)+2(2x-1)}{(2x+1)(2x-1)}\right)} \\ \\ \\ :\mapsto\tt{\:\displaystyle \lim_{x \to \frac{1}{2}} \left(2 \times \dfrac{(3x+2)\cancel{(2x-1)}}{\cancel{(2x-1)}(2x+1)}\right)} \\ \\ \\ \small\tt{\green{\:\:\:\:take\:limit\:at\:x\:to\:\frac{1}{2}}} \\ \\ \\ :\mapsto\tt{\:2 \times \dfrac{3\times \dfrac{1}{2}+2}{2 \times \dfrac{1}{2}+1}} \\ \\ \\ :\mapsto\tt{\:2 \times \dfrac{\dfrac{3}{2}+2}{1+1}} \\ \\ \\ :\mapsto\tt{\:2 \times \dfrac{\dfrac{3+4}{2}}{2}} \\ \\ \\ :\mapsto\tt{\:\cancel{2} \times \dfrac{7}{\cancel{2}}\times \dfrac{1}{2}} \\ \\ \\ :\mapsto\tt{\orange{\:\dfrac{7}{2}\:\:\:\:\:\:Ans}}$

$\large{\underline{\underline{\mathfrak{\tt{\blue{Important\:Formula}}}}}}$

$\bigstar\tt{\red{\:(a^2-b^2)\:=\:(a-b)(a+b)}} \\ \\ \bigstar\tt{\red{\:(a+b)(c+d)\:=\:(ac+ad+bc+bd)}}$