Evaluate the lim_x\to\4 of x-4\sqrtx^2-16

[wind]

Evaluate the limit

\(\displaystyle \L\mbox{\lim_{x\to\4} \frac{x-4}{\sqrt{x^{2}-16}\)

\(\displaystyle \L\mbox{\lim_{x\to\4} \frac{x-4}{\sqrt{x^{2}-16}\)*\(\displaystyle \L\frac{\sqrt{x^{2}+16}}{\sqrt{x^{2}+16}}\)

\(\displaystyle \L\mbox{\lim_{x\to\4}\frac{x-4\sqrt{x^{2}+16}}{x^{2}-16}\)

\(\displaystyle \L\mbox{\lim_{x\to\4}\frac{x-4\sqrt{x^{2}+16}}{(x+4)(x-4)}\)

\(\displaystyle \L\mbox{\lim_{x\to\4}\frac{\sqrt{x^{2}+16}}{x+4}\)

\(\displaystyle \L\frac{\sqrt{4^{2}+16}}{4+4}\)

\(\displaystyle \L\frac{1}{2}\)

the answer is supposed to be 0..what did I do wrong? Something in the second line?

Thanks

 

[Unco]

\(\displaystyle \mbox{ sqrt{x^2 - 16}sqrt{x^2 + 16} = \sqrt{(x^2 - 16)(x^2 + 16)}}\) (provided \(\displaystyle \mbox{x^2 - 16 \geq 0}\)), so indeed your second and third lines are a bit dodgy.

Instead, you might reconsider the expression as \(\displaystyle \mbox{ \sqrt{\frac{(x - 4)^2}{x^2 - 16}}\), etc.

 

[soroban]

Hello, wind!

Evaluate the limit: \(\displaystyle \L\L\lim_{x\to 4} \frac{x\,-\,4}{\sqrt{x^2\,-\,16}}\)

\(\displaystyle \L\lim_{x\to4} \frac{x\,-\,4}{\sqrt{x^2\,-\,16}}\,*\,\frac{\sqrt{x^2\,+\,16}}{\sqrt{x^2\,+\,16}}\;\;\) ??

\(\displaystyle \L\lim_{x\to4}\frac{(x\,-\,4)\sqrt{x^{2}\,+\,16}}{x^{2}\,-\,16}\;\;\;\) . . . no

Here's another approach . . . do some algegra first.

We have: \(\displaystyle \L\:\frac{x\,-\,4}{\sqrt{x^2\,-\,16}} \;=\;\frac{x\,-\,4}{\sqrt{(x\,-\,4)(x\,+\,4)}} \;=\;\frac{x\,-\,4}{\sqrt{x\,-\,4}\,\sqrt{x\,+\,4}} \;=\;\frac{\sqrt{x\,-\,4}}{\sqrt{x\,+\,4}}\)

Then: \(\displaystyle \L\:\lim_{x\to4}\,\frac{\sqrt{x\,-\,4}}{\sqrt{x\,+\,4}} \;=\;\frac{\sqrt{0}}{\sqrt{8}} \;=\;0\)


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Your method would have worked ... if you used the right conjugate.

\(\displaystyle \L\frac{x\,-\,4}{\sqrt{x^2\,-\,16}}\,*\,\frac{\sqrt{x^2\,-\,16}}{\sqrt{x^2\,-\,16}} \;=\;\frac{(x\,-\,4)\sqrt{x^2\,-\,16}}{x^2\,-\,16}\;=\;\frac{(\sout{x\,-\,4})\sqrt{x^2\,-\,16}}{(\sout{x\,-\,4})(x\,+\,4)}\;=\;\frac{\sqrt{x^2\,-\,16}}{x\,+\,4}\)


Then: \(\displaystyle \L\L\lim_{x\to4}\frac{\sqrt{x^2\,-\,16}}{x\,+\,4} \;=\;\frac{0}{8} \;=\;0\)