2

I'm having difficulties with this limit. We haven't covered integrals, derivatives, or things such as L'Hopital's rule in our course yet. What is the correct process to go about this limit?

$$\lim\limits_{x \to +\infty} \frac{\cos(2x^{-\frac{1}{2}})-1}{\frac{\pi}{2}-\arctan(3x)}$$

Larry
  • 5,090
Samuele B.
  • 761

2 Answers2

1

We have that by $\frac{\pi}{2}-\arctan(3x)=\arctan \frac1{3x}$ and standard limits

$$\frac{\cos(2x^-\frac{1}{2})-1}{\frac{\pi}{2}-\arctan(3x)}=\frac{\cos(2x^-\frac{1}{2})-1}{4x^{-1}}\frac{\frac1{3x}}{\arctan \frac1{3x}}\cdot12\to -\frac12\cdot 1 \cdot12=-6$$

user
  • 154,566
  • thank you! I'm afaid I don't understand the whole reasoning (it's me--I'm in need of lots of practice...)

    why does $\frac{\pi}{2} - \arctan(3x) = \arctan \frac{1}{3x}$ ?

    also, I'm trying to see how you got to that factorization. If I'm not making silly mistakes, then $2x^-\frac{1}{2} = \frac{1}{\sqrt{2x}}$ so how do I get from that to the factorization you wrote? Thank you and sorry if I'm asking obvious questions, I'm clearly a noob haha

     – Samuele B. Oct 13 '19 at 13:52
  • 1
    @SamueleB. For the identity refer to here. – user Oct 13 '19 at 13:57
  • @SamueleB. For the factorization the rule is to factor out expression in such way we obtain standard limits. In this case, since we have $\cos t-1$ at the numerator, we start with $\frac{\cos t -1}{t^2}$ and so on. – user Oct 13 '19 at 13:58
  • thank you so much! So $\arctan(3x) + \arctan(\frac{1}{3x}) = \frac{\pi}{2} \Leftrightarrow \frac{\pi}{2} - \arctan(3x) = \arctan \frac{1}{3x}$ makes total sense. Didn't know about the identity. – Samuele B. Oct 13 '19 at 14:02
  • @SamueleB. Note that it is true for $(3x)>0$. It is not so difficult to prove. – user Oct 13 '19 at 14:08
  • so I think I get it. The first fraction on the RHS makes total sense, although now that we've added $4x^-1$ something else needs to be added to keep it equivalent. $\frac{1}{3x}$ now gets added to the numerator of the second fraction so we get the standard limit of $-\frac{1}{2}$. But now we have multiplied all the expression by $\frac{1}{12x x^-1} = \frac{1}{12}$ which is why we finally multiply by $12$ and the equation is now verified. Correct? – Samuele B. Oct 13 '19 at 14:24
  • 1
    Yes exactly! Well done. – user Oct 13 '19 at 14:29
0

Set $\sqrt x=1/y$

and use Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function? to find

$$\lim_{y\to0^+}\dfrac{\cos2y-1}{\arctan\dfrac{y^2}3}$$

$$=-\lim\dfrac{2\sin^2y}{\cdots}$$

$$=\cdots=-\dfrac2{\dfrac13}$$

lab bhattacharjee
  • 274,582