0

$$\lim_{x\to\infty}|\sin(\pi\sqrt{x^{2} + x + 1})| $$

Couldn't do much as I did not get a clue on how to proceed.

Like the argument inside the square root has minimum value of $3/4$ and for different values of $x$ as we vary $x$ the square root part can become $5/4$ (then $\sin(5\pi/4) = 1/\sqrt{2}$) and of course for $x = 0 (\sin(\pi) =0)$ and for some $x$ we can also possibly find the answer as $\sin(\pi/2) = 1$.

Though when $x$ approaches infinity how to solve it?

Átila Correia
  • 15,530
Ramesh Sharma
  • 29
  • The limit does not exist because the function oscillates between 0 and 1. – DOUGLAS BRUNSON May 30 '22 at 01:19
  • This question appeared in a test,and the options were definite (a) 0 (b) 1/2 (c) 1/sqrt 2 (d) 1.And apparently the answer is 1 – Ramesh Sharma May 30 '22 at 01:21
  • OP, have you transcribed the problem correctly? Since $\sqrt{x^2+x+1}$ hits every natural number for some $x$ it's possible to find a sequence $x_n$ such that $\sin(\pi\sqrt{x_n^2+x_n+1}) = \sin(\pi n) = 0$. But it's also possible to find a sequence of numbers $y_n$ such that $\sqrt{y_n^2+y_n+1} = n+1/2$, in which case the function would be $1$. – Glare May 30 '22 at 01:24
  • @RamanujanXXV I'm having a look.Thanks. – Ramesh Sharma May 30 '22 at 01:31

0 Answers0