Solving Basic Limit Questions: When t=0, is it 0, -1, or Undefined?

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Homework Statement



<br /> \begin{align*}<br /> f(t) = \lim_{k \to \infty} f_k(t) = \lim_{k \to \infty} \frac{1 - kt^2}{1 +<br /> kt^2} = \lim_{k \to \infty} \frac{\frac{1}{k} - t^2}{\frac{1}{k} +<br /> t^2} = \frac{0 - t^2}{0 + t^2} = - \frac{t^2}{t^2}<br /> \end{align*}<br />

What is the value of limit function f when t = 0? Is it 0 or -1 or undefined? What is the reasoning behind it?

Does anyone know any good websites or books to catch up on these material?

Homework Equations





The Attempt at a Solution

 
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None of the above!

If t\ne 0 then the limit is -1, obviously.

If t= 0, go back to the original formula: if t= 0, then
\frac{1- kt}{1+ kt}= \frac{1- 0}{1+ 0}= \frac{1}{1}= 1
which is independent of k. The limit, if t= 0, is 1.
 
HallsofIvy said:
None of the above!

If t\ne 0 then the limit is -1, obviously.

If t= 0, go back to the original formula: if t= 0, then
\frac{1- kt}{1+ kt}= \frac{1- 0}{1+ 0}= \frac{1}{1}= 1
which is independent of k. The limit, if t= 0, is 1.

Thanks for your reply. I have one question about getting to the solution.

When should I use the original formula first and when should I take the limit first?
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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