Question

Solve
$lim_{x➝1} \frac{(2x-3)( \sqrt{1 } - x )}{ 2 {x}^{2} + x - 3 }$
​

Answer 1

$\huge{\underline\mathbf {\red{{Answer:}}}}$

$\mathbf{\frac{-1}{10}}$

Step-by-step explanation:

In this question,

We have to find the value of the given equation as x tends to 1

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{2x^{2}+x-3}$

$\textsf{This expression in the form} \frac{0}{0}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{2x^{2}+3x-2x-3}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{x(2x+3)-1(2x+3)}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{(x-1)(2x+3)}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{(\sqrt{x})^{2}-1^{2})(2x+3)}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)(2x+3)}$

$lim_{x\implies1}\frac{(2x-3)(\sqrt{x}-1)}{(\sqrt{x}+1)(2x+3)}$

Now putting the value of x as 1 We get,

$\frac{2\times 1 - 3}{(1+1)(2+3)} </p><p>$

$\mathbf{\frac{-1}{10}}$

$\huge{\underline\mathbf {\red{{HOPE \: \: \: IT \: \: \: \: \: HELPS}}}}$