Sufficiently rigorous limit proof?

Question

I needed to make the substitution $x = \sqrt{t^2+1} - t$ in an integral where in one limit $t\to+\infty$. So $x\to 0$ at this limit. I am aware you could complete the square under radical with an expression greater than $\sqrt{t^2+1} - t$, and then use the fact that this greater expression tends to zero. But my question is, is it simply not rigorous enough to say that as $t\to+\infty$., $t^2+1\to t^2$ so that $\sqrt{t^2+1}\to t$ and so $x\to 0$ ?

Sorry it took me a while to get to the actual question! Many thanks!

Note that $lim_{t \to \infty} t^2+1-t^2 = 1$, therefore $t^2+1 \to t$ as $t \to \infty$ is not true. — Hilario Fernandes, May 22 '15 at 21:41

Bernard · Accepted Answer · 2015-05-22T22:28:09.480

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No, it is not rigourous, and it can lead to false limis.

(Counter-)example:

You could use the same argument to saythat, as $\,t^2+t\to t^2$ as $t\to\infty$, $\sqrt{t^2+t}-t\to 0$.

Unfortunately, using the conjugate expressions, you'll find the limit is $\dfrac12$.

First method: conjugate expression: $$\sqrt{t^2+t}-t=\frac{(\sqrt{t^2+t})^2-t^2}{\sqrt{t^2+t}+t}=\frac t{\sqrt{t^2+t}+t}=\frac 1{\sqrt{1+\dfrac1t}+1}$$ (I supposed $t>0$)

Send method: binomial formula $$\sqrt{t^2+t}=t\sqrt{1+\dfrac1t}=t\Bigl(1+\dfrac1{2t}+o\Bigl(\dfrac1t\Bigr)\Bigr)=t+\dfrac12+o(1)$$ hence $$\sqrt{t^2+t}-t=\dfrac12+o(1)$$

edited May 22 '15 at 22:28

answered May 22 '15 at 21:41

Bernard

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I see. Sorry I know this wasn't in the original question, but how could one go about evaluating that limit? – user140591 May 22 '15 at 21:47
Do you mean the counter-example in my post? – Bernard May 22 '15 at 21:54
Yeah, the sqrt(t^2+t) - t as t -> 1/2. I'm unsure what you meant by using the conjugate expressions? – user140591 May 22 '15 at 21:58
I added some details. Hope this is clear, albeit succinct. – Bernard May 22 '15 at 22:12
I think you meant to write $(\sqrt{t^2+t})^2-t^2$ in the conjugate-expression method. – David K May 22 '15 at 22:25
Yes! Thank you for pointing the typo. – Bernard May 22 '15 at 22:28

score 0 · Answer 2 · answered May 23 '15 at 03:32

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Hint: Let $t=\sinh u$, as $u\to\infty$. Then use $\cosh u\pm\sinh u=e^{\large\pm u}$.

answered May 23 '15 at 03:32

Lucian

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score 0 · Answer 3 · answered May 23 '15 at 04:29

$$x = \sqrt{t^2+t} - t=t \sqrt{1+\frac 1 {t}}-t=t\Big(\sqrt{1+\frac 1 {t}}-1 \Big)$$ Now, remember that, for small $a$, $$\sqrt{1+a}=1+\frac{a}{2}-\frac{a^2}{8}+O\left(a^3\right)$$ Replace $a$ by $\frac 1{t}$ and get $$x=t\Big(1+\frac{1}{2 t}-\frac{1}{8 t^2}+\cdots-1)=\frac{1}{2}-\frac{1}{8 t}+\cdots$$

Sufficiently rigorous limit proof?

3 Answers3