Limit of $\cos\left(\frac{1}{(1-x)^2}\right)$ when $x=1$. I believe this function might be undefined at this point due to the $\frac{1}{0}$ which appears, however I'm not sure if this is correct and if so how to show that.
Asked
Active
Viewed 65 times
1 Answers
2
When we take the limit we are assuming that $x\neq 1$ then it doesn't matter if the function is not defined for $x=1$.
Take also a look here Why are we allowed to cancel fractions in limits?.
For the limit $\cos\left( \frac{1}{(1-x)^2} \right)$the problem is that $\frac{1}{(1-x)^2} \to \infty$ and $\lim_{t\to \infty} \cos t$ doesn't exist.
To show that let consider
- $t_n=2\pi n\to \infty \implies \cos (t_n)\to 1$
but
- $t_n=(2n+1) \pi\to \infty \implies \cos (t_n)\to -1$
user
- 154,566
-
He's right, limit is undefined in the interval (-1,1) – Karen May 15 '18 at 10:14
-
@Karen The problem is not that $1/(1-x)^2$ is undefined at $x=1$ the problem is that $\cos t$ has not limit for $t\to \infty$. – user May 15 '18 at 10:17
-
Should the final line be $(2n + 1)\pi$? – Adam Jones May 15 '18 at 10:21
-
@AdamJones opsss....thanks for pointing out that orrible typo! – user May 15 '18 at 10:22
-
I did not say that the problem is that $1/(1-x)^2$ is undefined, I said the WHOLE limit is undefined in the interval (-1,1), I gave you the reason. – Karen May 15 '18 at 10:24
-
@Karen the limit exist for any $x\to x_0$ with $x_0\neq 1$. What do you mean saying that the limit in udefined in $(-1,1)$? what is the limit as $x\to 0$? Isn't it $\cos 1$? – user May 15 '18 at 10:33
-
1@gimusi, I don't know if Karen would use these words/had a formal definition in mind, but I expect they meant that as $x $ approaches 1, the limit superior of the function is 1 and the limit inferior is -1. It's just a bit of extra info beyond "the limit doesn't exist". (IIRC, Mathematica would output the interval $[-1,1]$ if asked for the limit.) – Mark S. May 15 '18 at 10:39
-
@MarkS. Ah ok of course liminf and limsup are $-1$ and $1$! – user May 15 '18 at 13:30