Question

please give me solution for this problem​

Answer 1

$\rule{200}{2}$

$\Huge\bigstar\:\tt\underline\red{GIVEN}$

• $\:\:\Large\sf\blue{\lim_{x\to\frac{1}{2}}\left(\frac{f(x)-f(\frac{1}{2})}{2x-1}\right)}$

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$\Huge\bigstar\:\tt\underline\red{TO\:FIND}$

• $\:\:\sf\blue{Value\:of\:limit\:where\:f(x)=x^2+x-1}$

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$\Huge\bigstar\:\tt\underline\red{SOLUTION}$

$\huge\dashrightarrow\:\:\sf\blue{\lim_{x\to\frac{1}{2}}\left(\frac{f(x)-f(\frac{1}{2})}{2x-1}\right)}$

$\Large\leadsto\sf\purple{lim_{x\to\frac{1}{2}}\left(\frac{(x^2+x-1)-(\frac{1}{2})^2-\frac{1}{2}+1)}{2x-1}\right)}$

$\Large\leadsto\sf\green{\lim_{x\to\frac{1}{2}}\left(\frac{x^2+x-\frac{3}{4}}{2x-1}\right)}$

$\Large\leadsto\sf\purple{\lim_{x\to\frac{1}{2}}\left(\frac{(4x^2+4x-3}{4(2x-1)}\right)}$

$\Large\leadsto\sf\green{\lim_{x\to\frac{1}{2}}\left(\frac{(4x^2+6x-2x-3)}{4(2x-1)}\right)}$

$\Large\leadsto\sf\purple{\lim_{x\to\frac{1}{2}}\left(\frac{2x(2x+3)-1(2x+3)}{4(2x-1)}\right)}$

$\Large\leadsto\sf\green{\lim_{x\to\frac{1}{2}}\left(\frac{\cancel{(2x-1)}(2x+3)}{4\cancel{(2x-1)}}\right)}$

$\Large\leadsto\sf\purple{\lim_{x\to\frac{1}{2}}\left(\frac{(2x+3)}{4}\right)}$

$\Large\star\:\:\sf\underline\pink{Taking\:limit\:at\:x\:=\:\frac{1}{2} }$

$\huge\leadsto\sf\purple{\frac{(2\:×\:\frac{1}{2}+3)}{4}}$

$\huge\leadsto\sf\green{\frac{1+3}{4}}$

$\huge\leadsto\sf\purple{\cancel{\frac{4}{4}}}$

$\huge\leadsto\mathfrak\pink{1}$

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