Question

lim x tending to 1 (√x-1)(2x-3)/(2x²+x-3)

i hope u all will help me :)

Answer 1

$\large\underline{\sf{Solution-}}$

Given expression is

$\rm :\longmapsto\:\displaystyle\lim_{x \to 1}\rm \frac{( \sqrt{x} - 1) (2x - 3)}{ {2x}^{2} + x - 3 }$

can be further rewritten as

$\rm \:  =  \: \displaystyle\lim_{x \to 1}\rm (2x - 3) \times \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{ {2x}^{2} + x - 3}$

$\rm \:  =  \: (2(1) - 3) \times \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{ {2x}^{2} + 3x - 2x - 3}$

$\rm \:  =  \: (2 - 3) \times \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{ x(2x + 3) - 1(2x + 3)}$

$\rm \:  =  \: (- 1) \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{ (2x + 3) (x - 1)}$

$\rm \:  =  \: - \displaystyle\lim_{x \to 1}\rm \frac{1}{2x + 3} \times \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{ {( \sqrt{x} )}^{2} - {1}^{2} }$

$\rm \:  =  \: - \dfrac{1}{2 + 3} \times \displaystyle\lim_{x \to 1}\rm \frac{ \sqrt{x} - 1}{( \sqrt{x} - 1)( \sqrt{x} + 1) }$

$\rm \:  =  \: - \dfrac{1}{5} \times \displaystyle\lim_{x \to 1}\rm \frac{ 1}{\sqrt{x} + 1}$

$\rm \:  =  \: - \dfrac{1}{5} \times \dfrac{ 1}{\sqrt{1} + 1}$

$\rm \:  =  \: - \dfrac{1}{5} \times \dfrac{ 1}{1 + 1}$

$\rm \:  =  \: - \dfrac{1}{5} \times \dfrac{ 1}{2}$

$\rm \:  =  \: - \dfrac{1}{10}$

Hence,

$\rm\implies \:\boxed{\tt{ \displaystyle\lim_{x \to 1}\rm \frac{( \sqrt{x} - 1) (2x - 3)}{ {2x}^{2} + x - 3 } = - \frac{1}{10} \: }}$

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LEARN MORE

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{sinx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{tanx}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{log(1 + x)}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {e}^{x} - 1}{x} = 1 \: }}$

$\boxed{\tt{ \displaystyle\lim_{x \to 0}\rm \frac{ {a}^{x} - 1}{x} = loga\: }}$

Answer 2

Answer:

your solution is in above pic

pls mark it as brainliest