Please help I have this and I didn't get it well

Question

How to solve $\lim_{x \rightarrow 2} \frac{1}{x-2}$?

I tried to solve it by dividing the both sides on $x$ but it doesn't work and I tried to get the dam $-2$ out but also didnt work so how to solve it I know that there is no limit for this but how to Remove the indeterminacy

What indeterminacy do you need to remove? You are dividing by something that is approaching $0.$ That is usually sufficient. If you want to be very rigorous say $M$ is your upper bound. For an $M,$ you can find an $x$ in the neighborhood of $2$ such that $f(x) > M.$ Then show that there is also an x in the neighborhood such that $f(x) < - M$ — Doug M, Nov 08 '16 at 00:51
Not all limits that appear undefined can be made prettier to have an answer. — greenturtle3141, Nov 08 '16 at 00:54
Plotting a graph or checking a table of values always serve to give some light when you find a problem difficult to solve, as knowing the answer helps guide the way. — Simply Beautiful Art, Nov 08 '16 at 01:21
And welcome to MSE! We hope you truly enjoy and I welcome you be presenting the magic MathJax Handbook — Simply Beautiful Art, Nov 08 '16 at 01:22

score 0 · Answer 1 · answered Nov 08 '16 at 00:50

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You perform sign analysis with 1.9 and 2.1. If x = 1.9, then $\frac{1}{x-2}$ will be negative. If x = 2.1, then it will be positive. Thus, the limit approaching from the left is negative infinity and the limit approaching from the right is positive infinity, so the limit as x goes to 2 does not exist.

answered Nov 08 '16 at 00:50

Sapphira

286

1

This is... sort of true. However, think of something like $\sin(\frac{1}{x})$. Your analysis crashes and burns here – Brevan Ellefsen Nov 08 '16 at 01:22
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Thanks guys you for helping me – kkmm Nov 11 '16 at 00:25
@KazemJalabi Always welcomed. – Simply Beautiful Art Nov 11 '16 at 00:34

Simply Beautiful Art · Accepted Answer · 2016-11-08T15:04:50.330

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Actually, you should show the limit doesn't exist. Start by assuming they exist and that

$$\lim_{x\to2}\frac1{x-2}=\lim_{u\to0}\frac1u\tag{$x-2=u$}$$

Now, consider $u\to0^+$ and $u\to0^-$. Since we have

$$-\left(\frac1u\right)=\frac1{(-u)}$$

We must have

$$\underbrace{\lim_{u\to0^+}\frac1u}_L=\lim_{u\to0^-}\frac1{(-u)}=\underbrace{-\left(\lim_{u\to0^-}\frac1u\right)}_{-L}$$

Since, for the limit to exist, we must have the limit from the left equal the limit from the right, which in turn equal the limit from both sides, we end up with the following statement:

$$L=-L$$

where $L=\lim_{u\to0}\frac1u$. Solving this gives $L=0$, which isn't true since as $u\to0$, $\frac1u$ gets farther and farther from $0$. Thus, the limit cannot exist.

edited Nov 08 '16 at 15:04

answered Nov 08 '16 at 01:10

Simply Beautiful Art

74,685

Hm, does assuming the limit exists mean that the limit is equal to the average of it's left and right limits? – Simply Beautiful Art Nov 08 '16 at 01:27
It seems odd to start your argument by seeing that two "limits" that don't exist are equal. – Airymouse Nov 08 '16 at 01:48
@Airymouse Well, if you assume they exist, then they are equal. If one doesn't exist, then both don't exist. – Simply Beautiful Art Nov 08 '16 at 01:58
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There's a story that someone asked Bertrand Russell to assume that 1=2 and prove he was the Pope. Russell said,"the Pope and I are two, so the Pope and I are one." So you could just as well say, if you assume they exist, then they are unequal. What you are using is that if the limit on the right-hand side of your equation does not exist, then neither does the limit on the left-hand side. I'm guessing the questioner will have as much difficulty seeing this proposition as the one he queried. – Airymouse Nov 08 '16 at 02:31

Please help I have this and I didn't get it well

2 Answers2