Question

Hi guys! I really need help with this limits question.. :/

Ok, so the Question is to find the limit as x --> 16 in the function (((square root of x) -4)/(x-16)).

I tried plugging in the 16, but got 0/0 (indeterminate). So, I'm trying to use the conjugate trick, but I'm getting a little lost. Can anyone help?

Answer 1

First, let me verify that I have this function correct:\[(((\square \root of x) -4)/(x-16))\]

Answer 2

Unfortunately, I'm unsure what you mean by "(square root of x) -4)" It would be so much better if you could present this expression symbolically. You could draw it in the Draw utility, below, or use the Equation Editor (also below). Or you could share a screen shot.

Answer 3

I tried to take a screenshot of the problem! Please tell me if it works correctly, if not, I can draw it!

Answer 4

\[\lim_{x \rightarrow 16} \frac{ \sqrt{x}-4 }{ x -16 }\]

Answer 5

I figured out how to write it on the equation writer!

Answer 6

You still there?

Answer 7

@ashvad

Answer 8

\[\lim_{x \rightarrow 16} \frac{ \sqrt{x}-4 }{ x -16 }\]

Answer 9

does indeed produce 0/0 if we substitute 16 for x. This result is undefined.

One approach here would be to factor the denominator as you would factor the difference between two squares. x becomes \[(\sqrt{x})^2\]

Answer 10

and so the denominator is then \[(\sqrt{x}-2)(\sqrt{x}+2)\]

Answer 11

\[(\sqrt{x}-4)(\sqrt{x}+4)\]

Answer 12

Note my error, above; the factors involving 4 are correct, whereas those involving 2 are not.

Answer 13

Now reduce the following:\[\frac{ \sqrt{x}-4 }{( \sqrt{x}-4 )(\sqrt{x}+4)}\]

Answer 14

Finally, let x approach 16. Your result?

Answer 15

Ohh!! I understand now! So once I do the difference of two squares factoring, I can cancel out the numerator, leaving a 1 and then when I plug in 16 for x, I end up getting 1/8 as my answer! (:

Answer 16

@mathmale Thank you so much!

Answer 17

1/8 sounds just right!

Answer 18

My pleasure!