Question

The value of the limit $\underset{x\to \infty }{lim}(\sqrt{x^2+x}-x)$ is equal to

• A. $\underset{x\to \infty }{lim}\sqrt{x^2+x+1}-\sqrt{x^2+1}$
• B. $\underset{x\to -\infty }{lim}\frac{x}{x-\sqrt{x^2-2x}}$
• C. $\underset{x\to 0}{lim}(\cos 2x{)}^{\frac{\log (\sqrt{2}-x)}{x^2{\cos }^{2}x}}$
• D. $\underset{x\to -1}{lim}\left(\frac{x+1}{4}\right)\tan (-\frac{\pi }{2}x)$

Answer 1

Answer: (A), (B), (C)

$\underset{x\to 0}{lim}(\cos 2x{)}^{\frac{\log (\sqrt{2}-x)}{x^2{\cos }^{2}x}}$

$\underset{x\to \infty }{lim}(\sqrt{x^2+x}-x)=\underset{x\to \infty }{lim}\frac{x}{(\sqrt{x^2+x}+x)}=\underset{x\to \infty }{lim}\frac{1}{(\sqrt{1+\frac{1}{x}}+1)}=\frac{1}{2}$

$\underset{x\to \infty }{lim}\sqrt{x^2+x+1}-\sqrt{x^2+1}=\underset{x\to \infty }{lim}\frac{x}{\sqrt{x^2+x+1}+\sqrt{x^2+1}}=\underset{x\to \infty }{lim}\frac{x}{x\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}+x\sqrt{1+\frac{1}{x^2}}}=\frac{1}{2}$

$\underset{x\to \infty }{lim}\frac{x}{x+\sqrt{x^2-2x}}$
$=\underset{x\to \infty }{lim}\frac{x}{x+x\sqrt{1-\frac{2}{x}}}=\underset{x\to \infty }{lim}\frac{1}{1+\sqrt{1-\frac{2}{x}}}=\frac{1}{2}$

$\underset{x\to 0}{lim}(\cos 2x{)}^{\frac{\log (\sqrt{2}-x)}{x^2{\cos }^{2}x}}=e^{\underset{x\to 0}{lim}\frac{\log (\sqrt{2}-x)}{x^2{\cos }^{2}x}(-2{\sin }^{2}x)}=e^{(-2)\log \left(\sqrt{2}\right)}=(\sqrt{2}{)}^{-2}=\frac{1}{2}$

$\underset{x\to -1}{lim}\left(\frac{x+1}{4}\right)\tan (-\frac{\pi }{2}x)=\underset{x\to -1}{lim}\left(\frac{t}{4}\right)\tan (-\frac{\pi }{2}(t-1\left)\right)$
By replacing $x+1=t$ such that $x\to -1\Rightarrow t\to 0$ So, we have
$\frac{1}{4}\underset{t\to 0}{lim}\frac{\frac{\pi }{2}\left(t\right)}{\tan \frac{\pi }{2}\left(t\right)}\frac{2}{\pi }=\frac{1}{4}\cdot \frac{2}{\pi }=\frac{1}{2\pi }$