Is My Initial Approach to Proving a Function's Limit Correct?

  • Thread starter Thread starter Questions999
  • Start date Start date
  • Tags Tags
    Function Limit
Questions999
Messages
151
Reaction score
0
Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε

I know how to keep this going,but is it correct until now?
 
Physics news on Phys.org
Elaia06 said:
Prove that the limit when x--> infinite of [(-1 )^n-1]/n^2=0
So for ε > 0,exists N>0 so that n>N => |x -a|< ε
The right side should be |[(-1 )^(n-1)]/n^2| < ε
Elaia06 said:
What I do is |[(-1 )^n-1]/n^2| < ε. Here I remove the absolute value and I have (1^n-1)/n^2 < ε
|(-1)^(n - 1)| = 1, so all you need to do is find n so that 1/n^2 < ε.
Elaia06 said:
I know how to keep this going,but is it correct until now?

Put parentheses around your exponent. What you wrote, (-1)^n - 1 would be interpreted as
$$(-1)^n - 1 $$
 
Thank you so much! :)
 
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...

Similar threads

Back
Top