Question

plss give me solution for this??? asap​

Answer 1

$\large{\underline{\underline{\mathfrak{\green{\sf{Solution:-}}}}}}$.

$\leadsto\:1$

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$\large{\underline{\underline{\mathfrak{\sf{Given\:Here:-}}}}}$.

• $\large\boxed{\displaystyle\lim_{x\to\frac{1}{2}}\left(\frac{f(x)-f(\frac{1}{2}}{2x-1}\right)}$

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$\large{\underline{\underline{\mathfrak{\sf{Find\:Here:-}}}}}$.

• $\:value\:of\:limit\:where\:f(x)\:=\:x^2+x-1$

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$\large{\underline{\underline{\mathfrak{\sf{Explanation:-}}}}}$.

$\large\boxed{\displaystyle\lim_{x\to\frac{1}{2}}\left(\frac{f(x)-f(\frac{1}{2}}{2x-1}\right)}$

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

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

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

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

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

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

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

$\bold{\:taking\:limit\:at\:x\:=\frac{1}{2}}$

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

$\leadsto\frac{1+3}{4}$

$\leadsto\frac{4}{4}$

$\leadsto\:1$

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Answer 2

Step-by-step explanation:

arey abhi Tumne answer correct hi Diya hai phir kyu mujse pooch rahai ho