Simplify and guess the value of $\lim_{x\to 2}\frac {x^3-8}{x-2}$.
I'm very confused on how to do this. Please help!
$\endgroup$ 15 Answers
$\begingroup$Quick way: We know that $x^3 - 8 = x^3 - 2^3$, so this is a difference of cubes, which factors into $(x - 2)(x^2 + 2x + 2^2)$. So your expression is the same as $\frac{(x-2)(x^2 + 2x + 4)}{x - 2}$. Cancel the $x - 2$s. We get $x^2 + 2x + 4$. At $x = 2$, this is $12$, which is our answer.
But why does that work? Why are we allowed to cancel?
The trick is: we can cancel whenever $x - 2 \ne 0$. So we can cancel, with a restriction: $\frac{(x-2)(x^2 + 2x + 4)}{x - 2} = x^2 + 2x + 4$, for all $x \ne 2$. Conveniently, when we look at the limit of our function, we look at values of $x$ that are close to, but not equal to $2$. So we can conclude:
$$\lim_{x \to 2} \frac{(x-2)(x^2 + 2x + 4)}{x - 2} = \lim_{x \to 2} x^2 + 2x + 4$$
We know that polynomials are continuous, so the limit at any $c$ is the same as the value at $c$, so we just plug in $2$ and get $12$ as our answer. We don't have to go through this process every time, normally we just "cancel", but it is good to know that there's no trickery going on.
(Also, for math notation on SE, see here.)
$\endgroup$ $\begingroup$Provided you know some basic differential calculus, which may not be the case as you are asking a question about limits which is necessarily an earlier topic, here is an alternative method.
Letting $f(x) = x^3$, we have
$$\lim_{x \to 2}\frac{x^3 - 8}{x - 2} = \lim_{x \to 2}\frac{f(x) - f(2)}{x -2} = f'(2).$$
As $f'(x) = 3x^2$, the limit is $f'(2) = 12$.
$\endgroup$ $\begingroup$One way to do this is to approach this like the velocity question from before. You want to see what happens as $x$ gets closer and closer to $2$, so just check some nearby values. We say $$f(x)=\frac{x^3-8}{x-2}$$ And when $x$ gets closer to $2$, $f(x)$ gets closer to some limit. Plugging in a few values, we have: $$ f(3) = \frac{3^3-8}{3-2}=19\\ f(2.1)= 12.61\\ f(2.01)=12.0601\\ f(2.001)=12.006 $$ What value is $f(x)$ getting closer to as $x\to2$?
Alternatively, you can notice that for any $x$ not equal to $2$, we have $$ f(x)=\frac{x^3-8}{x-2}=\frac{(x-2)(x^2+2x+4)}{x-2}=x^2+2x+4 $$ Why does this help?
$\endgroup$ 3 $\begingroup$For $\lim\limits_{x\to 2}\dfrac{x^3−8}{x−2}$, when $x \to 2$, the result tends to positive or negative infinity.
Apply L'Hôpital's rule, we differentiate both denominator and numerator separately, we will get $\dfrac{3x^2}{1} = 3x^2$.
Put $x=2$, $3 \times 2^2 = 3\times 4 = 12$.
$\endgroup$ 2 $\begingroup$The above answers are not the solution you are looking for. When you are asked to "guess" a limit, you are to construct a table and test input values that approach 2. To be complete you should consider both values greater than 2 and values less than 2.
x 1.8 1.9 1.95 1.99 Und. 2.01 2.05 2.10 2.20 f(x) 10.84 11.41 11.70 11.94 ? 12.06 12.30 12.61 13.24
Now look for the trend. As x->2, f(x)-> 12. Since the output values are approaching 12, we can reasonably "guess" that the limit is equal to 12.
(NOTE: in a later section of your course you will be asked to "evaluate" the limit. That is what others were trying to explain, and unfortunately probably caused you even more confusion.)
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