How do you find $\lim _ { x \rightarrow 7} \frac { x ^ { 3} - 343} { x - 7}$ ?

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Answer 1 · 2018-01-27T03:36:59

$147$

Direct substitution does not work because it gives us 0/0, an indeterminate form.

We could try factorising this rational function so that we can possibly cancel out common terms.

The factorised numerator is: $(x-7)(x^2+7x+49)$
The denominator cannot be factorised further.

Therefore, we find that $lim_(xrarr7)(x^3-343)/(x-7)=lim_(xrarr7)((x-7)(x^2+7x+49))/(x-7)$ .

We can cancel the $(x-7)$ !

We get $lim_(xrarr7)x^2+7x+49$

If we substitute $x=7$ into $x^2+7x+49$ , we get:

$=7^2+7(7)+49$
$=49+49+49$
$=147$

This is the graph below.
graph{(x^3-343)/(x-7) [-24.6, 25.38, 129.5, 154.49]}

How do you find \lim _ { x \rightarrow 7} \frac { x ^ { 3} - 343} { x - 7}\lim _ { x \rightarrow 7} \frac { x ^ { 3} - 343} { x - 7}?