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When evaluating limits we often come across the form 0/0 and we cant get a limiting value. As stated in many textbooks and online resources, the method to still get a limiting value as the input approaches a specific value is to first factorize or simplify the function to cancel out the factors/variables responsible for the form 0/0.

My question is doesn't it change the function altogether if we simplify it?

Because if we had used the original function it would not have been defined at that particular input? Can anyone provide the intuition or concept behind this?

Batwayne
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    It does change the expression at the point under consideration. However, a limit does not care about the value of the expression at the point; it only matters what happens as you approach that point. As long as the expression does not change for points in the neighborhood, excluding the point itself, there is no issue. – Paul Aug 08 '16 at 12:36
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    @Merkh Not quite correct. The function has been changed because $\frac{x}{x}$ actually does not exist for $x=0$. But since $\frac{x}{x}=1$ for all $x\ne 0$, the limit exists and is $1$ because we just add one point to the original function. – Peter Aug 08 '16 at 12:42
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    @Merkh That is wrong. You can only conclude that $\frac{x}{x}=1$ if $x\neq 0$. A function with the formula $\frac{x}{x}$ for all points in its domain must have a domain that excludes $x=0$, whereas a function with formula $1$ can have domain $\mathbb{R}$. – smcc Aug 08 '16 at 12:42
  • I like to remember this simply as "Removable Singularities Don't Affect Limits" (at least in the real plane.... It's a bit more complicated in the complex plane, but more or less carries over with additional constraints) :) – Brevan Ellefsen Aug 08 '16 at 14:52
  • see http://math.stackexchange.com/a/1822706/72031 – Paramanand Singh Aug 09 '16 at 11:04

2 Answers2

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The main thing is:

Taking the limit does not mean evaluating the expression at the point, but rather infinitely close to that point (i.e. approaching the point).

Simplifying an expression does change the behaviour at the point (for example getting a value where it previously was indeterminate because of division by 0 or something else), but it does not change the behaviour close to the point, i.e. as you approach the point, which is what matters for the limit.

naslundx
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Let examine an example;

$$\lim_{x\to1} (\frac{x-1}{x^2 -1}) =\lim_{x\to1} \frac{x-1}{(x-1)(x+1)} = \frac{1}{2}$$

Since x is not 1 but around 1, we can do the simplification.

if we didn't do this, the function wouldn't be defined at 1 but it doesn't matter because we are not interested in the value of the function.

And it doesn't chance the function in matter value except at the point the function is undefined

Our
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