0
$\begingroup$

Here is a problem about limit tends to infinity.I have given an answer.Please let me know,if this is correct

  • 0
    What is the question? – 2017-02-19
  • 0
    I want to check if my answer is correct. – 2017-02-19
  • 0
    https://i.stack.imgur.com/BPanf.jpg – 2017-02-19
  • 2
    Your answer is correct, but I believe you mean $1+3^{-\infty}$ in the denominator in second last line. – 2017-02-19
  • 0
    yeah,I think so. – 2017-02-19
  • 0
    I dk why someone voted down. I think down votes should require written reasons and be subject to review. – 2017-02-19
  • 0
    Note that limits don't tend to anything. They're not moving around doing the cha-cha-cha. They either exist or don't. – 2017-02-19

2 Answers 2

0

The answer is correct. Just write $$\lim_{n\to \infty} \frac{3^n - 3^{-n}}{3^n + 3^{-n}}= \lim_{n\to \infty} \frac{3^n}{3^n}\cdot \frac{1 - 3^{-2n}}{1 + 3^{-2n}} = 1$$ and add that $\lim_{n\to \infty} 3^{-2n} =0$. I personally would avoid to write something like $3^{-\infty}$.

0

Yes your answer is correct. Any integer divided by zero is infinity except zero.

Either it's 1/0 or -2/0.