Question

Evaluate the limit as x approaches 1 of the quotient of the square root of the quantity x squared plus 3 minus 2 and the quantity x minus

Answer 1

\[\lim_{x \rightarrow 1} \frac{\sqrt{x^2+3}-2}{x}\]?

Answer 2

x-1 in the denominator, but yes that is right

Answer 3

\[\lim_{x \rightarrow 1} \frac{\sqrt{x^2+3}-2}{x-1}\]

Answer 4

ok well pluggin in 1 on top gives us 0
and pluggin in 1 on bottom gives us 0
since we have 0/0 we have more work

Answer 5

I would definitely try rationalizing the numerator here

Answer 6

would you simply just square everything on top

Answer 7

you can't do that that is changing the value of a fraction
for example:
\[\frac{3}{1} \neq \frac{3^2}{1^2}\]


to rationalize something you might want to think about the conjugate of the thing you want to rationalize

Answer 8

That is multiply the top and bottom by top's conjugate

Answer 9

\[\frac{\sqrt{x^2+3}-2}{x-1} \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}=?\]
just multiply the top out for now

Answer 10

Finding the limit algebraically sometimes calls for one or some of the following (and this is list is not a list of everything):
A) rationalizing
B) factoring
C) combining fractions
D) substitution
E) Sometimes I referring to other limits helps also.

Answer 11

okay so would the top cancel out and you be left with 1/(x-)((square root of x^2 +3)+2)

Answer 12

(a-b)(a+b)=a^2-b^2 <--- so it shouldn't give you 1

Answer 13

Im so confused

Answer 14

\[(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)=(\sqrt{x^2+3})^2-2^2=?\]

Answer 15

x^2+9-4

Answer 16

which eqauls x^2+5 do you set it equal to zero then and solve for x

Answer 17

\[(\sqrt{x^2+3}-2)(\sqrt{x^2+3}+2)=(\sqrt{x^2+3})^2-2^2=x^2+3-4\]

Answer 18

More simply put, recall from algebra that\[(a+b)(a-b)=a^2-b^2\]\[a=\sqrt{x^2+3} \]\[b=2\]

Answer 19

So\[x^2+3-4\]is\[x^2-1\]

Answer 20

\[\frac{\sqrt{x^2+3}-2}{x-1} \cdot \frac{\sqrt{x^2+3}+2}{\sqrt{x^2+3}+2}=\frac{x^2+3-4}{(x-1)(\sqrt{x^2+3}+2)}\]

Answer 21

now you should be able to see the top is factorable

Answer 22

and then that the top and bottom have a common factor

Answer 23

do i have to multiply the two functions together in the denominater

Answer 24

no

Answer 25

just factor the top

Answer 26

and cancel a common factor on top and bottom

Answer 27

\[\frac{x^2-1}{(x-1)(\sqrt{x^2+3}+2}=\frac{(x+1)(x-1)}{(x-1)(\sqrt{x^2+3}+2}\]

Answer 28

would my x-1 cancel then

Answer 29

yes

Answer 30

then would i use one as x to find the limit

Answer 31

\[\frac{x+1}{\sqrt{x^2+3}+2}\]
now you can plug in 1

Answer 32

\[\frac{x-1}{x-1}\frac{(x+1)}{\sqrt{x^2+3}+2}=\frac{x+1}{\sqrt{x^2+3}+2}\]

Answer 33

then that will give me my limit?

Answer 34

yep