Proofs wether a limit exists or not.

Question

Let me define a function:-

$$f(x)=\begin{cases} 2x & \text{, } x≠5 \\ x & \text{, } x=5 \end{cases}$$

Now let me find:-

$$\lim_{x \to\ 5} f(x) = 10$$

Now I will try to prove that this limit exists.

According to the formal definition of limits:-

$$ 0<|x-c|< \delta \implies 0<|f(x)-L|< \epsilon$$

I would not mention here the other requirements and conditions because I think people here know the formal definition of limits.

So that would mean we should be able to express $\delta$ in terms of $\epsilon$.

Let me try to do this:-

Using the definition:-

$$0<|x-5|<\delta$$

To express it in the form of $\epsilon$. We will try to get the above inequality in the form of $|f(x)-L|< \epsilon$. :-

$$0<2×|x-5|< 2× \delta$$

Which is the same as :-

$$0<|2x-10| < 2\delta \text{, } x≠5$$

It doesn't matter that $x≠5$ as in limits we don't talk about what will happen when we reach there.

As we see now our L.H.S. is in the form $|f(x)-L|$ but our R.H.S. is different. we can then equate the R.H.S. to express it in terms of $\epsilon$.

$$ 2\delta= \epsilon \Longleftrightarrow \delta = \frac{\epsilon}{2} $$

I think that the above thing does not makes sense.

for example $4<5$ and $4<6$ but $5≠6$.

So how can one say that $2\delta=\epsilon$ base on just this fact that $|f(x)-L|$ is less than both as we see 4 is less than both 5 and 6 but this does not mean that 5=6.

Answer 1

The core of the proof is this:

If $0<|x-5|<\delta$, and if $\delta = \frac{\epsilon}{2}$, then $|f(x)-10| < \epsilon$.

The statement above is true for all values of $\epsilon>0$, which means $10$ is indeed the limit of $f$ as $x$ goes to $5$.

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Everything else in the proof is just a particular method of how to get the exact values of $\delta$ and $\epsilon$.

There is no such thing in the proof as saying $4<5$ and $4<6$, therefore $5=6$. The proof is more similar to saying "I need some value $a$ that would satisfy the inequality $4<a$. I also know that the inequality $4<5$ is true, therefore, I can say that $a=5$ satisfies the inequality".

And this sort of reasoning is perfectly OK if you want to prove a statement of the type "There exists some $a$ for which the inequality $4<a$ is satisfied.