Question

solve question one pls fast​

Answer 1

Answer:

1/4

Step-by-step explanation:

We need to evaluate the given limit:

$\sf \lim\limits_{x\to0}\dfrac{-2+\sqrt{4+x}}{x^2 + x}$

$\sf\implies \lim\limits_{x\to0}\dfrac{\sqrt{4+x} - 2}{x^2 + x}$

${\sf\implies \lim\limits_{x\to0}\dfrac{\sqrt{4+x} - 2}{x^2 + x}\cdot\dfrac{\sqrt{4+x} + 2}{\sqrt{4+x}+2}}$

${\sf\implies \lim\limits_{x\to0}\dfrac{(\sqrt{4+x} - 2)(\sqrt{4+x} + 2)}{(x^2 + x)(\sqrt{4+x} + 2)}}$

${\sf\implies \lim\limits_{x\to0}\dfrac{(\sqrt{4+x})^2 - (2)^2}{(x^2 + x)(\sqrt{4+x} + 2)}}$

${\sf\implies \lim\limits_{x\to0}\dfrac{4 + x - 4}{x(x + 1)(\sqrt{4+x} + 2)}}$

${\sf\implies \lim\limits_{x\to0}\dfrac{ x }{x(x + 1)(\sqrt{4+x} + 2)}}$

${\sf\implies \lim\limits_{x\to0}\dfrac{1}{(x + 1)(\sqrt{4+x} + 2)}}$

${\sf\implies \dfrac{1}{(0 + 1)(\sqrt{4+0} + 2)}}$

${\sf\implies \dfrac{1}{1(2+ 2)}}$

${\sf\implies \dfrac{1}{4}}$

Answer 2

Step-by-step explanation:

$\color{magenta} \underline{ \begin{array}{ || |l| || } \hline \color{magenta} \\ \hline \boxed{ \text{ \tt \: Solution:-} } \end{array}}$

$\color{darkviolet} \pmb{\begin{aligned} \tt \lim _{x \rightarrow-\infty}\left(\sqrt{x^{2}-x+1}+x\right) & \tt=\lim _{y \rightarrow \infty}\left(\sqrt{y^{2}+y+1}-y\right), \text { where } y=-x \\ \\ & \tt=\lim _{y \rightarrow \infty} \frac{\left\{\sqrt{y^{2}+y+1}-y\right\}\left\{\sqrt{y^{2}+y+1}+y\right\}}{\left\{\sqrt{y^{2}+y+1}+y\right\}} \\ \\ & \tt=\lim _{y \rightarrow \infty} \frac{y^{2}+y+1-y^{2}}{\sqrt{y^{2}+y+1}+y} \\ \\ & \tt=\lim _{y \rightarrow \infty} \frac{y+1}{ \sqrt{ {y}^{2} + y + 1 } + y} \\ \\ & \tt =\lim _{y \rightarrow \infty} \frac{1+\frac{1}{y}}{1+\frac{1}{y}+\frac{1}{y^{2}}+1}=\frac{1}{2}\end{aligned} }$