Question

Another tricky limits question :/

Hi guys! I've encountered another tricky indeterminate limits question. Could anyone please walk me through how to do this??

Answer 1

I assume you do not want to use L'Hospital ?

if not, then try multiplying top and bottom by the "conjugate" of the top
i.e. by \[ \sqrt{x^2+3}+2\]

Answer 2

I would vote for your using phi's latest suggestion. Make certain that you get the algebra right. If you do, then you need only substitute -1 for x in your result; this should give you the desired limit immediately.

Answer 3

I tried using the conjugate, but I think my math is wrong somewhere. I'm having a bit of difficulty with the radicals. When I multiplied with the conjugate, I got\[\frac{ x ^{2} + 3 - 4 }{ x + 1 \sqrt{x^2 +3} +2}\]

Is that step correct? If not, where did I go wrong, and if yes, where do I go from here? Because I tried to plug in -1 at this step, but I still get 0/0 or indeterminate.

Answer 4

Hey Ash :)
When running into these types of problems,
`only expand out the conjugate multiplication`.

Leave the bottom two pieces sitting next to each other.

Answer 5

Oh I guess you did :O
You meant to write this,\[\large\rm \frac{ x ^{2} + 3 - 4 }{ (x + 1)(\sqrt{x^2 +3} +2)}\]yes?

Answer 6

Simplify the numerator,
pssst, it's the difference of squares, it will factor.

Answer 7

When I factored out the numerator, I got \[\left( x-1 \right) \left( x+4 \right) \]

that doesn't seem right though, because it would seem that we would want to cancel out that x+1. Did I do something wrong?

Answer 8

and yes! That is what I meant earlier with the conjugate! (:

Answer 9

Nooo silly :O

\(\large\rm x^2+3x-4=(x-1)(x+4)\)
But we don't have a 3x in the middle.

Combine the 3 and 4,\[\large\rm x^2+3-4=x^2-1\]ya?

Answer 10

Ohhh, I definitely need to brush up on that! Thank you!
So okay, can I then factor that out to
\[\left( x + 1 \right)\left( x - 1 \right)\]
then cancel the \[\left( x+ 1 \right)\]

in the numerator and denominator?

Leaving me with

\[\frac{ \left( x-1 \right) }{ \sqrt{x^2 +3}+2 }\]
which when I plug in '-1' gives me
\[\frac{ \left( (-1) -1 \right) }{ \sqrt{(-1)^2 + 3}+2 }\]

= \[\frac{ -2 }{ 4 }\]

??

Answer 11

yayyyy good job \c:/

Answer 12

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