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I am trying to prove this using two different cases. One case shows the limit "$l$" equals some number leads to contradiction and the other one shows that the limit "$l$" is not equal to that same number also leads to contradiction.

I am able to do the first case.

Case 1: let $l=1$. Let $\epsilon=0$. Let $\delta$>0.

Let m be a natural number so that $m > \frac{1}{\delta}$. Then $0<\frac{1}{m}<\delta$.

Let $x=\frac{1}{m}$. This means $0 < x < \delta$. Also, $f(x)=m^2$ and since $m$ is a natural number, $m^2$ is greater than or equal to $1$.

So $0<|x-0|< \delta$ and yet $|\frac{1}{x^2}-1| = |f(x) - l| \geq 0$. So if $l=1$, this means $1$ is not the limit as $x$ goes to zero of $f(x)$.

For case $2$ where $l\neq 0$, I have a possible solution, but I think I may have let certain numbers depend on other numbers in a way that does not work.

Case 2: $l$ is not $1$. Let $\epsilon=-l$. Let $\delta>0$. Let $m$ be a natural number so that $\frac{1}{\delta} < m < \sqrt{2l}$.

Then $1 < \delta m < \delta\sqrt{2l}$. So $0 < \frac{1}{m} < \delta < \delta\sqrt{2l}$. (I know that $1/m$ is greater than zero since m is a natural number).

Let $x = 1/m$. Now we have $0 < x < \delta$. $f(x) = m^2$. $f(x) < 2l$ since $m < \sqrt{2l}$.

Notice $2l = l - \epsilon$ and also that $0 = l + \epsilon$ So, since $m>0$, we have $0 < f(x) < 2l$ which gives us $l + \epsilon < f(x) < l - \epsilon$. Rearranging this inequality I can get $|f(x) - l|> \epsilon$

So if $l$ is not equal to zero, $l$ is not the limit $f(x)$ as $x$ goes to zero

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HINT: Trying to show that $\lim_{x\rightarrow 0} \frac{1}{x^2} \neq 0$ gives you a contradiction will not work because that statement is true. Rather what you should try to show that $\lim_{x\rightarrow 0} \frac{1}{x^2} = L$ for $L \in \mathbb{R}$ always leads to a contradiction.

Edit: Also, note that $m^2 \rightarrow \infty$ as $m \rightarrow \infty$.

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  • $\begingroup$ I just made an edit to my post, does the argument I make not work for the reason that you have said? $\endgroup$ – mmm Oct 10 '14 at 20:46
  • $\begingroup$ You could make it work that way but I do not think you need a special case for $L = 0$. Also, I do not think your case 2 works, since you can choose $\delta = \frac{1}{2} = L$ so that $\frac{1}{\delta} = 2 > 1 = \sqrt{2L}$, which contradicts the inequality you use. Try just choosing $m$ so that $1/m$ is 'large enough'. $\endgroup$ – Bruce Zheng Oct 11 '14 at 15:48
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It is easy to work in the following way:

Suppose $1/x^2\to L$ as $x\to 0$. By the limit laws, $x^2\cdot 1/x^2\to 0\cdot L=0$, as $x\to 0$, but $(x^2\cdot 1/x^2)=1\to 1$ and by uniqueness of the limit $1=0$, contradiction.

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