Question

Evaluate the limit:

Answer 1

\[\lim_{x \rightarrow 3^{+}}\frac{ \sqrt{x ^{2}-9} }{ x-3}\]

Answer 2

\[\lim_{x \rightarrow 3^{+}}\frac{ \sqrt{x ^{2}-9} }{ x-3 }*\frac{ \sqrt{x ^{2}+9} }{ \sqrt{x ^{2}+9} } = \frac{ x ^{2}-9 }{ x-3(\sqrt{x ^{2}-9)} }\]

that's right?

Answer 3

You should have \(\sqrt{x^2+9}\) in the denominator. Then factor the numerator and the \(x-3\) factors disappear.

Answer 4

\[\lim_{x \rightarrow 3^{+}}\frac{ \sqrt{x ^{2}-9} }{ x-3 }*\frac{ \sqrt{x ^{2}+9} }{ \sqrt{x ^{2}+9} } = \frac{ x ^{2}-9 }{ x-3(\sqrt{x ^{2}-9)} }\]
So true solution?

Answer 5

\[\lim_{x \rightarrow 3^{+}}\frac{ \sqrt{x ^{2}-9} }{ x-3 }*\frac{ \sqrt{x ^{2}+9} }{ \sqrt{x ^{2}+9} } = \frac{ x ^{2}-9 }{ x-3(\sqrt{x ^{2}-9)} } = \frac{ (x+3)(x-3) }{x-3 (\sqrt{x ^{2}-9)} }= \frac{ (x+3) }{ \sqrt{x ^{2-9}} }\]

Answer 6

Right, but it should be
\[\lim_{x\to3^+}\frac{x+3}{\sqrt{x^2+9}}\]
You can plug in 3 right away.

Answer 7

thank you :)

Answer 8

yw

Answer 9

Hold on a sec, I skimmed over your work again and I think something might be wrong with it...

Answer 10

Like what?

Answer 11

The main issue I'm seeing is that
\[\sqrt{x^2-9}\cdot\sqrt{x^2+9}\not=x^2-9\]

Answer 12

so what now ? :(
I think it's true ..

Answer 13

It's not, you're confusing it with
\[\sqrt{x^2-9}\cdot\sqrt{x^2-9}=x^2-9\]

Answer 14

Distributing, we'd get
\[\sqrt{x^2-9}\cdot\sqrt{x^2+9}\not=\sqrt{x^4-81}\]
which I don't think is very helpful...

Answer 15

so ?

Answer 16

\[\sqrt{x ^{2}-9} \times \sqrt{x ^{2}+9} = (\sqrt{x ^{2}-9)^{2}} = (x ^{2}-9)\]

Answer 17

Because the square root will go with quadrature

Answer 18

You're saying that
\[(x^2-9)(x^2+9)=(x^2-9)^2\]
which is not the case.

Instead of rewriting the expression algebraically to evaluate the limit, you could instead plug in numbers increasingly closer to \(x=3\), coming from the right. In other words, evaluate the expression for \(x=3.1,~3.01,~3.001\) and so on.

Answer 19

this is my solution

Answer 20

The problem still lies in the step between the first two lines. The algebra doesn't work out like that.

Also, the limit is not finite.

Answer 21

Could it be that the problem of the question may be what is wrong in it?

Answer 22

I'm saying that there's no way to algebraically manipulate the expression to allow you to remove the discontinuity at \(x=3\).

Answer 23

limit does not exist (or +infinity in this case)

since x -> 3+, (x-3) is always positive so we can do this

sqrt(x^2-9) / sqrt( (x-3)^2 )

and x^2-9 is also positive, so we can do this

sqrt[ (x^2-9) / (x-3)^2 ]

sqrt[ (x+3) / (x-3) ]

when evaluating the limit, it's positive infinity

Answer 24

hmmm yes, in the end you'd need to evaluate by checking a few values for "x"

Answer 25

\(\bf \large{\cfrac{\sqrt{x^2-3^2}}{x-3}\implies \cfrac{\sqrt{(x-3)(x+3)}}{x-3}\implies
\cfrac{(x-3)^{\frac{1}{2}}(x+3)^{\frac{1}{2}}}{(x-3)^1}
\\ \quad \\
(x-3)^{\frac{1}{2}}\cdot (x-3)^{-1}(x+3)^{\frac{1}{2}}\implies (x-3)^{\frac{1}{2}-1}(x+3)^{\frac{1}{2}}
\\ \quad \\
(x-3)^{-\frac{1}{2}}(x+3)^{\frac{1}{2}}\implies \cfrac{\sqrt{x+3}}{\sqrt{x-3}}}\)

but pretty much what sourwing said