I am trying to show that $\lim_{t \to \infty} \cos t + t^2 $ does not exist. Without usint deltas and epsilons, I would argue by contradiction. if not, then $L = \lim_{t \to \infty} \cos t + t^2$ where $L$ may be infinity. Then we would have

$$ \lim_{t \to \infty} \cos t = L - \lim_{t \to \infty} t^2 $$

Which would be me a contradiction since we $\lim \cos t$ Does not exist.

What I think is that this argument looks like of fishy. What if $L = \infty$? then we would have $\infty - \infty $ on the right, which is undetermined form.

What other argument for calculus students can we use to use the nonexistence of such limit?

  • $\begingroup$ The values cos takes will be between -1 to 1, so yes L will also be $\infty$, can't you just say that it doesn't exist because $t^2$ diverges $\endgroup$ – Nikunj Feb 25 '16 at 10:46
  • $\begingroup$ Please parenthesize. $\endgroup$ – Yves Daoust Feb 25 '16 at 11:00
  • $\begingroup$ $\lim_\limits{t \to \infty} \big(f(t) + g(t)\big) = \Big(\lim_\limits{t \to \infty} f(t)\Big) + \Big(\lim_\limits{t \to \infty} g(t)\Big)$ This result is true if both $\lim_\limits{t \to \infty} f(t)$ and $\lim_\limits{t \to \infty} g(t)$ exists. $\endgroup$ – user297008 Feb 25 '16 at 11:14

Take any $\;R\in\Bbb R^+\;$ , so there exists $\,M\in\ R\;$ such that $\;x>M\implies x^2>R+1\;$, and then for for these same $\;x>M\;$ :

$$\cos x+x^2\ge-1+x^2>R\implies\lim_{x\to\infty}\left(\cos x+x^2\right)=\infty$$

and the limit exists though it is not finite.

  • 1
    $\begingroup$ Please, if anyone downvotes tell me what you found incorrect. It serves the asker and also me. Thanks. $\endgroup$ – DonAntonio Feb 25 '16 at 11:02

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