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Consider the limit $$\lim_{x\rightarrow 1}\frac {x^2-1}{x-1} $$ Plugging in $1 $ for $ x $ will yield $\frac00 $, which is no good. However, if we simplify the expression, we get: $$\lim_{x\rightarrow 1}\frac {x^2-1}{x-1} = \lim_{x\rightarrow 1}(x+1)$$ Now, plugging in $1 $ yields the answer $2 $. My question is: why on earth should simplifying the expression help us find the limit? The two expressions above are absolutely equivalent. I.e., if I plug in, say $10 $ in $(x+3)^2 $ and $x^2+6x+9 $, I get the same answer - $169 $.

Mauro ALLEGRANZA
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user132181
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    The two expressions are not completely equivalent. The first expression has a singularity at $x=1$, the second does not. The clever thing is that both expressions are equivalent on all points except $x=1$, but by the definition of limit, the actual value at $x=1$ is irrelevant. That's why the trick works. – a06e May 23 '14 at 14:44
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    To be clear, $\lim_{x\rightarrow1}\frac{x^2-1}{x-1}$ does indeed evaluate to $2$. It is $\frac{x^2-1}{x-1}$ which is undefined at $x=1$. – Pockets May 23 '14 at 14:48
  • As @becko said, the first equation is not defined for $x=1$. When you are taking the limit, you look at numbers that are arbitrarily close to $1$, so therefore you are justified in cancelling out the terms in the numerator and denominator. – The very fluffy Panda May 23 '14 at 14:48

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They are not absolutely equivalent, although in this case they're equal everywhere except at $x=1$. Thus formally, as functions, they're different.

It is this close relationship between the two functions that allows this lemma:

If functions $f,g$ are equal on some interval $(a,b)$ except possibly at a point $c\in(a,b)$, and if $\lim_{x\to c}f(x)$ exists, then so does $\lim_{x\to c}g(x)$, and the two limits are equal.

This is very easy to prove even with an epsilon delta argument. The key fact is that you can exchange $g(y)$ with $f(y)$ for any $y$ other than $x$ in the interval.

So the moral of the story is that limits can't tell the difference between two functions that differ only at a point (and clearly the same is true for any set of points which can be trapped away from each other by intervals.)

rschwieb
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You can only simplify the expression when $x\neq 1$. In all other cases ($x\neq 1$) the expression is $x+1$. Hence you simply write $x+1$ for all $x\in \mathbb R -\text{{1}}$ and then take the value as you approach $1$ from both sides, which in this case is $2$.

Apurv
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  • The first sentence definitely addresses a misconception of the poster, but the second doesn't really explain anything, it just asserts that it is true.... Maybe some additions are in order? – rschwieb May 23 '14 at 15:02
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    @rschwieb, I don't call rewriting the expression as manipulation. It is natural. No one canexplain why it helps. – Apurv May 23 '14 at 15:04
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Limits tell us about the behaviour of a function in a neighbourhood, and not at the mentioned point.

In your example $$\frac{x^2-1}{x-1} and\ x+1 $$ are the same things, except at the point $x=1$. So, the two representations are equivalent as long as you don't go to $x=1$.

Now, the limit at $x=1$ will tell us what value the function approaches(not necessarily the exact value of the function at the point). Since, in the vicinity of $x=1$ the function behaves exactly as $x+1$, therefore we substitute $x+1$ for the function $\frac {x^2-1}{x-1}$.

Shubham
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First, you need to ask yourself why plugging a number in the expression works? Precisely speaking, you need to figure out the condition under which the following expression holds $$\lim_{x\to a}f(x)=f(a)$$

The answer is the continuity. In other words, for a continuous function $f\colon\Bbb R\to\Bbb R$, we have $$\lim_{x\to a}f(x)=f(a)\quad\forall a\in\Bbb R$$

Now, is $g(x)=\frac{x^2-1}{x-1}$ continuous in $\Bbb R$? Obviously not, because $g$ is not even defined on $x=1$. Therefore, plugging a number will never works for $g$.

Second, why simplifying a expression works? The answer is very simple: you can always simplify a expression as long as you can. Besides, limit of a function at $x=1$ is only a behavior of a function in the neighborhood of $x=1$

Therefore, to calculate the limit $$\lim_{x\to 1}\frac{x^2-1}{x-1}$$ we first note that $(x^2-1)/(x-1)$ is not continuous at $x=1$. Therefore, plugging $x=1$ in the expression will not work. Then, in the neighborhood of $x=1$, $g$ is equivalent to $(x+1)$, we simplify the expression and get $$\lim_{x\to 1}(x+1)$$ Now, is $(x+1)$ continuous at $x=1$? Of course, it is! Now, you can plug $x=1$ in $(x+1)$ to get the final answer.

YYF
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