Check existence of limit with definition

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SUMMARY

The discussion revolves around verifying the limit $\lim_{x\to 0}\frac{x}{x}$ using the epsilon-delta definition of limits. The user confirms that to demonstrate this limit, one must show that for every $\epsilon>0$, there exists a $\delta>0$ such that if $0<|x-0|<\delta$, then $|f(x)-1|<\epsilon$. The conclusion reached is that since $x \neq 0$, the expression simplifies to zero, confirming the limit is indeed 1.

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mathmari
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Hey! :o

I want to check the existence of the limit $\lim_{x\to 0}\frac{x}{x} $ using the definition.

For that do we use the epsilon delta definition?

If yes, I have done the following:

Let $\epsilon>0$. We want to show that there is a $\delta>0$ s.t. if $0<|x-0|<\delta$ then $|f(x)-1|<\epsilon$.

We have that $\left |f(x)-1\right |=\left |\frac{x}{x}-1\right |=\left |\frac{x-1}{x}\right |=\frac{|x-1|}{|x|}$.

How can we continue? (Wondering)
 
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Hey mathmari!

Since $0<|x-0|$, it follows that $x\ne 0$.
Therefore $\left|\frac xx -1\right|=0$, isn't it? (Wondering)
 
Klaas van Aarsen said:
Since $0<|x-0|$, it follows that $x\ne 0$.
Therefore $\left|\frac xx -1\right|=0$, isn't it? (Wondering)

Ohh yes! Thank you! (Blush)
 

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