Question

Calculus

Answer 1

only #52

Answer 2

factor the denominator

Answer 3

\(x-9 = x - 3^2\)

Answer 4

rationalise the denominator @Lena772

Answer 5

\((\sqrt{x})^2 - 3^2\)

Answer 6

you should pull an identity now..

Answer 7

(x-3)(x-3)

Answer 8

careful

Answer 9

\(a^2-b^2 = (a+b)(a-b)\)

Answer 10

the sqrt is only over the x in the numerator

Answer 11

yes,

denominator we have just \(x-9\)

Answer 12

\(x\) can be written as \((\sqrt{x}) ^2\)

Answer 13

9 can be written as \(3^2\)

Answer 14

so that we can use the known identity

Answer 15

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

Answer 16

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-9}\)

Answer 17

\[\frac{ 2\sqrt{x}-6 }{ (x-9)}\]

Answer 18

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-3^2}\)

Answer 19

fine ?

Answer 20

yep

Answer 21

apply the identity on denominator

Answer 22

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-3^2}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x}+3)(\sqrt{x}-3)}\)

Answer 23

The last step i got lost

Answer 24

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-3^2}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x}+3)(\sqrt{x}-3)}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\)

Answer 25

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{x-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-9}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x})^2-3^2}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2\sqrt{x}-6}{(\sqrt{x}+3)(\sqrt{x}-3)}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}\)

\(\large \lim \limits_{x \to 9} ~~~\frac{2}{(\sqrt{x}+3)}\)

Answer 26

now you can take the limit
plug x = 9

Answer 27

2/6

Answer 28

1/3

Answer 29

\(\large \color{red}{\checkmark}\)