Lim of x=4 (x^(3/2) – 8) / (x-4)

[Sophie]

Lim of x=4 (x^(3/2) – 8) / (x-4)

Hello I have to use appropriate factorization to find the limit of:
Lim of x=4 (x^(3/2) – 8) / (x-4)

I have used the values of 4.1, 4.01, 4.001 to guess at a limit of 3.

However I cannot factor out the limit. My initial response was to try and get a value of (x-4) in the numorator, but I did not get very far with that. I am convinced the denominator will be (Sqx + 2) (Sqx –2), I am a little lost here…

If someone could point me in the right direction I would be grateful.

Thanks Sophie

 

[pka]

Multiply \(\displaystyle \L \frac{{\sqrt {x^3 } - 8}}{{x - 4}}\left( {\frac{{\sqrt {x^3 } + 8}}{{\sqrt {x^3 } + 8}}} \right)\) out.

Then factor \(\displaystyle x^3 - 64\) as the difference of two cubes.

 

[Sophie]

Thanks

Thanks

 

[soroban]

Re: Lim of x=4 (x^(3/2) – 8) / (x-4)

Hello, Sophie!

I have to use appropriate factorization to find the limit of:
. . \(\displaystyle \L\lim_{x\to4}\,\frac{x^{\frac{3}{2}}\,-\,8}{x\,-\,4}\)

The factoring is very difficultto see . . .

The numerator is a difference of cubes: \(\displaystyle \,(\sqrt{x})^3 - (2)^3\)
. . which factors: \(\displaystyle \\sqrt{x}\,-\,2)(x\,+\,2\sqrt{x}\,+\,4)\)

The denominator is a difference of squares: \(\displaystyle \,(\sqrt{x})^2\,-\,(2)^2\)
. . which factors: \(\displaystyle \\sqrt{x}\,-\,2)(\sqrt{x}\,+\,2)\)

The function becomes: \(\displaystyle \L\:\frac{(\sqrt{x}\,-\,2)(x\,+\,2\sqrt{x}\,+\,4)}{(\sqrt{x}\,-\,2)(\sqrt{x}\,+\,2)} \;=\;\frac{x\,+\,2\sqrt{x}\,+\,4}{\sqrt{x}\,+\,2}\)

Therefore: \(\displaystyle \L\:\lim_{x\to4}\,\frac{x\,+\,2\sqrt{x}\,+\,4}{\sqrt{x}\,+\.2}\;=\;\frac{4\,+\,2\cdot2\,+\,4}{2\,+\,2}\;=\;\frac{12}{4}\;=\;3\)