Help with Pre-Calc.?

What happens to the denominator as you approach 7 from the left?

x is slightly smaller than 7, so x - 7 will be a very small *negative* number, close to zero.

Then we square it, so it will be a very small *positive* number close to zero.

But we are taking the reciprocal of that (1/(x - 7)²) so the result will get bigger and bigger as we approach 7.

So it approaches ∞ from the left

What happens to the denominator as you approach 7 from the right?

x is slightly bigger than 7, so x - 7 will be a very small *positive* number, close to zero.

Then we square it, it will still be a very small *positive* number close to zero.

But we are taking the reciprocal of that (1/(x - 7)²) so the result will get bigger and bigger as we approach 7.

So it approaches ∞ from the right

Either way, the limit is ∞.

If you graph it, you'll see that there is a vertical asymptote at x = 7, as you would expect.

See the graph below.

Here, YOU must do it. You dont ask your neighbor to go and jog for you in the hopes it makes you get into shape, right? Same thing with math. YOU must practice a few times on your own, else you will never get good at it.

So, make a table (math) and compute many points (couples) for (x, f(x)).

Then use this table to make a graph of the function (sketch it or plot in on graph paper). .

Then, to confirm your drawing, plot it in a graphing calculator ( just type the function into google; it will automatically graph it for you).

Then, you can "see" the asymptotes & limits.

lim x->7 1/(x-7)^2 = +oo ( or is not defined )