# $\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$ without using L’Hôpital.

Can I calculate the following limit without applying L'Hôpital?

$$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)$$

With L'Hôpital it gives me as a result $\frac{-9}{2}$.

• How did you apply L'Hopital rule here? – Jack Oct 1 '17 at 14:19
• You've mistyped the problem. – zhw. Oct 1 '17 at 16:52

I think he want to type and solve this $$\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\frac{-9}{2}$$ this can be solve by multiplying conjugate $$\quad{\lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)=\\ \lim_{x\to\infty} x^2(cos(\frac{3x+2}{x^2})-1)\times \frac{cos(\frac{3x+2}{x^2})+1}{cos(\frac{3x+2}{x^2})+1}=\\ \lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\}$$ as $x\to \infty \implies \sin(\frac{3x+2}{x^2})\to 0 \implies \sin(\frac{3x+2}{x^2})\sim \frac{3x+2}{x^2}\\$so $$\quad{\lim_{x\to\infty} x^2(-\sin^2(\frac{3x+2}{x^2}))\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\ \lim_{x\to\infty} -x^2(\frac{3x+2}{x^2})^2\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\ \lim_{x\to\infty} -(\frac{(3x+2)^2}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\\lim_{x\to\infty} -(\frac{9x^2+12x+4}{x^2})\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\ \lim_{x\to\infty} -(\frac92)\times \frac{1}{cos(\frac{3x+2}{x^2})+1}=\\ -(\frac92)\times \frac{1}{cos(0)+1}=\\\frac{-9}{2}}$$

• @trick15f :please look at my answer !>? – Khosrotash Oct 1 '17 at 15:37
• @MariosGretsas : I saw the question , I saw the answer $\frac{-9}{2}$ and then think ,may be this – Khosrotash Oct 1 '17 at 16:28
• @Khosrotash thank you very much, I understood everything – trick15f Oct 1 '17 at 16:28
• @trick15f : please explain How you do a mistake in writing question ! – Khosrotash Oct 1 '17 at 16:29

Since $$\lim_{x\to\infty} \operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)=\cos1>0,$$ we obtain

$$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-1\right)=+\infty$$

Maybe do you mean $$\lim_{x\to\infty} x^2\operatorname{cos}\left(\frac{3x+2}{x^2}-\frac{\pi}{2}\right)=\infty?$$

• You used that $f(x)=\cos x$ is an even and continuous function. $$\lim_{x\to \infty}\left(\frac{3x+2}{x^2}-1\right)=$$ $$=\lim_{x\to \infty}\left(\frac{\frac{3}{x}+\frac{2}{x^2}}{1}-1\right)=$$ $$=\frac{0+0}{1}-1=0-1=-1$$ and $\cos (-1)=\cos 1$. – user236182 Oct 1 '17 at 14:36
• @user236182 Yes of course! – Michael Rozenberg Oct 1 '17 at 14:42

We have to find the limit of :

$$\lim\limits_{x \to \infty}x^2(cos(\frac{ax+b}{x^2})-1)=?$$

If we put the substitution $x=\frac{1}{y}$ and use the formula $1-2sin(x)^2=cos(2x)$ we get : $$\lim\limits_{y \to 0}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2}$$ Wich equivalent to : $$\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2\frac{(ay+by^2)^2}{4}}$$ Or $$\lim\limits_{y \to 0}\frac{-sin(\frac{(ay+by^2)}{2})^2}{\frac{(ay+by^2)^2}{4}}=-1$$ Because : $\lim\limits_{z \to 0}\frac{sin(z)}{z}=1$

So we have :

$$\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2sin(\frac{(ay+by^2)}{2})^2}{y^2\frac{(ay+by^2)^2}{4}}=\lim\limits_{y \to 0}\frac{(ay+by^2)^2}{4}\frac{-2}{y^2}=\frac{-a^2}{2}$$