Math, asked by tokaians, 1 year ago

evaluate the following limit Lt x tends to 1 (2x-3)(√x-1)/(2x^2+x-3)

Answers

Answered by Anonymous

124

see attachment hope it will help u ^_^

Attachments:

tokaians: limits is really very hard. thank you

Anonymous: no its preety cool and basic =_=

tokaians: it isn't for me

tokaians: each sum has a different logic ..............

Anonymous: yaa sometimes

tokaians: my exams are round the corner and I am not able to aolve this chapter properly

Anonymous: have u learnt basics of all intermediate forms

tokaians: yes

Answered by ujalasingh385

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

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

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,

$\frac{2\times 1 - 3}{(1+1)(2+3)}$

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

Previous Question

Next Question