limit $\frac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n}$ as $t_n-s_n \to + \infty$, where $t_n > s_n, \forall n.$

Question

I am trying to show that $$\frac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n} \to 1$$ $$\frac{t_n-s_n+\arctan(t_n)-\arctan(s_n)}{t_n-s_n} \to 1$$ as $t_n-s_n \to + \infty$, where $t_n > s_n, \forall n.$ But I have no clue how to do that since there is no information about $t_n$ and $s_n$ individually.
Any help would be great.

Answer 1

If $t_n-s_n \to \infty$, Find these values:

$\dfrac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n}$.

$\dfrac{t_n-s_n+\arctan(t_n)-\arctan(s_n)}{t_n-s_n}$.

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If $t_n$ and $s_n$ both converges, $t_n-s_n \not\to \infty$.

Therefore, $t_n$ or $s_n$ doensn't converge.

If $t_n$ doesn't converge:

$\dfrac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n} = 1.$

If $s_n$ doesn't converge:

$\dfrac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n} = 1.$

If $t_n $ and $s_n$ don't converge:

$\dfrac{\sqrt{t_n^2+1}-\sqrt{s_n^2+1}}{t_n-s_n}=\dfrac{t_n+s_n}{\sqrt{t_n^2+1}+\sqrt{s_n^2+1}}.$

So, if $t_n=-s_n+\alpha$, the value goes to $0$.

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$\dfrac{t_n-s_n+\arctan(t_n)-\arctan(s_n)}{t_n-s_n}=1+\dfrac{\arctan(t_n)-\arctan(s_n)}{t_n-s_n}.$

$-1 \leq \arctan(x) \leq 1.$

$\Rightarrow -2 \leq \arctan(t_n)-\arctan(s_n) \leq 2, \dfrac{\arctan(t_n)-\arctan(s_n)}{t_n-s_n}=0.$

$\therefore \dfrac{t_n-s_n+\arctan(t_n)-\arctan(s_n)}{t_n-s_n}=1.$