lim _{x rightarrow 1} frac{(2 x-3)(sqrt{x}-1) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$ Factorize the denominator: $2x^2+x-3 = (2x+3)(x-1)$ Substitute this i

Vedclass · Accepted Answer

$-\frac{1}{10}$

Vedclass · Answer

(B) Given limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3}$ Factorize the denominator: $2x^2+x-3 = (2x+3)(x-1)$ Substitute this into the limit: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{(2 x+3)(x-1)}$ Multiply the numerator and denominator by $(\sqrt{x}+1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)(\sqrt{x}+1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$ Simplify using $(\sqrt{x}-1)(\sqrt{x}+1) = (x-1)$: $\lim _{x \rightarrow 1} \frac{(2 x-3)(x-1)}{(2 x+3)(x-1)(\sqrt{x}+1)}$ Cancel $(x-1)$ from numerator and denominator: $\lim _{x \rightarrow 1} \frac{2 x-3}{(2 x+3)(\sqrt{x}+1)}$ Evaluate the limit by substituting $x=1$: $\frac{2(1)-3}{(2(1)+3)(\sqrt{1}+1)} = \frac{-1}{5(2)} = -\frac{1}{10}$

$\lim _{x \rightarrow 1} \frac{(2 x-3)(\sqrt{x}-1)}{2 x^2+x-3} = $

Similar Questions

$\lim _{x \rightarrow-\infty} \frac{5 x^3-x^2 \sin 5 x}{x \cos 4 x+7|x|^3-4|x|+3} = $

Evaluate the limit: $\lim_{x \rightarrow 0} \left( \frac{4^x - 1}{2^x - 1} - \frac{\sqrt{4 + 3x} - 2}{x} \right)$

If $f(x) = \begin{cases} \frac{\sin([x])}{[x]}, & \text{when } [x] \neq 0 \\ 0, & \text{when } [x] = 0 \end{cases}$ where $[x]$ is the greatest integer function,then $\lim_{x \to 0} f(x) = $

$\lim _{x \rightarrow \infty}\left(\frac{x+5}{x+2}\right)^{x+3}$ equals

$\mathop {\lim }\limits_{x \to 0} \frac{x}{|x| + {x^2}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module