1
$\begingroup$

I have problems to solve this limit $$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$ I tried with Taylor: $$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac{1}{2x}+\frac{1}{24x^2}\right)\right)=\lim _{x\to \infty }\left(x^2-x^2+\frac{x}{2}-\frac{1}{24}\right)=\infty$$ that is the wrong result, in must be $\color{red}{-\frac{1}{6}}$

$\endgroup$

3 Answers 3

3
$\begingroup$

Hint: You need to use the Taylor series expansion for the square root as well as the cosine. Expanding $\sqrt{1-u}$ around $u=0$ we have that $$\sqrt{1-u}=1-\frac{u}{2}-\frac{u^2}{8}+O(u^3)$$ and so as $x\rightarrow \infty$ $$x^2\sqrt{1-\frac{1}{x}}=x^2-\frac{x}{2}-\frac{1}{8}+O\left(\frac{1}{x}\right).$$

$\endgroup$
1
$\begingroup$

Full solution:

$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)= \lim _{x\to \infty }x^2\left(\sqrt{1-\frac{1}{x}}-\cos\left(\frac{1}{\sqrt{x}}\right)\right)=$$$$=\lim _{t\to 0}\frac{\sqrt{1-t^2}-\cos{t}}{t^4} =\lim _{t\to 0}\frac{1-t^2-\cos^2{t}}{t^4(\sqrt{1-t^2}+\cos{t})}=\frac{1}{2}\lim _{t\to 0}\frac{\sin^2t-t^2}{t^4}=$$$$=\frac{1}{2}\lim _{t\to 0}\frac{\sin t+t}{t}\cdot\lim _{t\to 0}\frac{\sin t-t}{t^3}=\frac{1}{2}\cdot2\cdot(-\frac{1}{6})=-\frac{1}{6}.$$

For the last limit to apply the rule of L'Hospital.

$\endgroup$
0
$\begingroup$

Another idea: $$ \lim_{x\to\infty}\left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right) = \lim_{x\to\infty}\frac{ \left(x\sqrt{x^2-x}-x^2\cos\left(\frac1{\sqrt{x}}\right)\right)\left(x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)\right)} {x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} = $$ $$ \lim_{x\to\infty}\frac{ x^2(x^2-x)-x^4\cos^2\left(\frac1{\sqrt{x}}\right)}{x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} = \lim_{x\to\infty}\frac{ -x^3+x^4\sin^2\left(\frac1{\sqrt{x}}\right)} {x\sqrt{x^2-x}+x^2\cos\left(\frac1{\sqrt{x}}\right)} = $$ $$ \lim_{x\to\infty}\frac{ -x+x^2\sin^2\left(\frac1{\sqrt{x}}\right)} {\sqrt{1-1/x}+\cos\left(\frac1{\sqrt{x}}\right)} = \cdots $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .