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I am having trouble with understanding limits. For example:

$\displaystyle \lim_{x\rightarrow 3}\;\frac{(x-3)(x^2-x-2)}{(x-3)} = 4$

(sorry for this notation but I'm new here) I understand that we can cancel (x-3) in numerator and denominator, because x will be close to 3 but never actually 3. But I don't understand why can you plug 3 after getting new function, in other words why will

$\displaystyle \lim_{x\rightarrow 3}\;(x^2-x-2) = 4$

have the same value as $3\cdot 3 - 3 - 2$ when we assumed that $x$ will never be $3$ but only close to $3$. So, should we instead plug $2.999$ and get limit $2.999\cdot2.999 - 2.999 - 2$?

Lehs
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user2648841
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3 Answers3

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Because $f(x)=\frac{(x-3)(x^2-x-2)}{x-3}$ and $g(x)=x^2-x-2$ agree at all values of $x$ except $x=3$, we can say that $$\lim_{x\to 3}\frac{(x-3)(x^2-x-2)}{x-3}=\lim_{x\to 3}(x^2-x-2)$$.

However, the latter function is continuous (its a polynomial), and so its limit can be evaluated by plugging in. But this also gives the value of the former limit, since the functions agree for $x\ne 3$.

paw88789
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Since when doing the limit $\;x\to 3\;$ we do not take $\;x=3\;$ but only values arbitrarily close to $\;3\;$ ,we can reduce:

$$\frac{\color{red}{(x-3)}(x^2-x-2)}{\color{red}{x-3}}=x^2-x-2\xrightarrow[x\to\ 3]{}9-3-2=4$$

Timbuc
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How close does $x$ get to 3? Is $2.999$ close enough? What about $2.9999$? $2.9999999999$? The basic idea when you take this limit is that $x$ gets so close to $3$ that it's virtually indistinguishable from $3$. As a result, in this case the limit becomes virtually indistinguishable from just plugging $3$ into the equation.

Scott Caldwell
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