Question

Find limit f(x)=The root of (x+3) - 1 / x+2 when x is heading towards -2

Answer 1

Alright, so you understand when you plugged in -2 for x, you got the form 0/0. correct?

Answer 2

yea

Answer 3

Usually when you have square roots in a limit problem, you'll have to multiply the fraction by the conjugate of the square root term.

Answer 4

For example, the conjugate of 'sqrt(x) - 2' is 'sqrt(x) + 2'

Answer 5

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

Answer 6

yea

Answer 7

Right, that's the function, we have a 0/0 form. Multiply both the numerator and the denominator by the conjugate of the square root term.

Answer 8

u mean we multiple it with \[\sqrt{x+3} +1\div \sqrt{x+3} +1\]

Answer 9

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

Answer 10

Oh yeah, you got it. Sorry, I didn't read your post right at first glance lol.

Answer 11

Hint: You don't have to expand the denominator usually.

Answer 12

So i get this eventually \[\frac{ 1 }{ \sqrt{x+3} +1 }\] right ?

Answer 13

Err, I'm not sure. Did you simplify the product?

Answer 14

and eventually it becomes \[\frac{ 1 }{ 2 }\]

Answer 15

Yea

Answer 16

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

Answer 17

It shouldn't be reduced to 1/2 because of the (x+2) binomial in the denominator.

Answer 18

I.k But x+3-1 = (x+2)

Answer 19

we take out the x+2
that leavs us with \[\frac{ 1 }{ \sqrt{x+3}+1 }\]

Answer 20

OH! Lol, correct, totally did not see that. It's early in the morning. So very sorry.
Yes, now the problem has become the limit as x approaches -2 of the new function!

Answer 21

And yes, it'll be 1/2.

Answer 22

Hahahaha N.p Thanks btw