Find the Limit

[Guest]

what is the limit to this question? Im not sure how you find the limit for this!

lim (x is approaching 3) (7-x)square rooted - 2 / [1- (4-x)square rooted]

umm im not sure if I said that right.. I meant square rooted as the symbol that looks like a divided sign with a tic at the end.

 

[stapel]

Do you mean the following?


. . . . .\(\displaystyle \large{\begin{array}{c}limit\\x\rightarrow 3\end{array}\,\, \sqrt{7\,-\,x}\,\,-\,\,\frac{2}{1\,-\,\sqrt{4\,-\,x}}}\)


If so, I would suggest rationalizing the denominator and then evaluating at x = 3.

If not, then please review the formatting articles in the "Forum Help" pull-down menu at the very top of the page, and reply with clarification.

Thank you.

Eliz.

 

[Guest]

sry I dont see the "forum help" but the -2 is part of the numerator and yes you have the denominator correctly written

 

[soroban]

Hello, bittersweet!

\(\displaystyle \L\lim_{x\to3}\,\frac{\sqrt{7\,-\,x}\,-\,2}{1\,-\,\sqrt{4\,-\,x}}\)

We must rationalize the numerator and the denominator . . .

\(\displaystyle \L\frac{\sqrt{7\,-\,x}\,-\,2} {1\,-\,\sqrt{4\,-\,x}} \,\cdot\,\frac{\sqrt{7\,-\,x}\,+\,2}{\sqrt{7\,-\,x}\,+\,2}\.\cdot\,\frac{1\,+\,\sqrt{4\,-\,x}}{1\,+\,\sqrt{4\,-\,x}} \;=\;\frac{[(7\,-\,x)\,-\,4](1\,+\,\sqrt{4\,-\,x})}{[1\,-\,(4\,-\,x)](\sqrt{7\,-\,x}\,+\,2)\)

\(\displaystyle \L\;\;=\;\frac{(3\,-\,x)(1\,+\,\sqrt{4\,-\,x})}{(x\,-\,3)(\sqrt{7\,-\,x}\,+\,2)} \;=\;\frac{-(\sout{x\,-\,3})(1\,+\,\sqrt{4\,-\,x})}{(\sout{x\,-\,3})(\sqrt{7\,-\,x}\.+\,2)} \;= \;-\frac{1\,+\,\sqrt{4\,-\,x}}{\sqrt{7\,-\,x}\,+\,2}\)

Therefore: \(\displaystyle \L\;\lim_{x\to3}\,\left[-\frac{1\,+\,\sqrt{4\,-\,x}}{\sqrt{7\,-\,x}\,+\,2}\right] \;=\;-\left(\frac{1\,+\,\sqrt{1}}{\sqrt{4}\,+\,2}\right)\;=\;-\frac{2}{4}\;=\;-\frac{1}{2}\)

 

[stapel]

sry I dont see the "forum help"

Scroll all the way up to the very top of the page. "Forum Help" is above the "FreeMathHelp" logo, right next to "Forum Index" and "Math Subjects". These are also all visible when you have your browser open to the "Post a reply" box.

Eliz.

 

[galactus]

Because it is of indeterminate form, \(\displaystyle \L\\\frac{0}{0}\), we could

also use L'Hopital's rule:

Take the derivative of the numerator:

\(\displaystyle \L\\\frac{d}{dx}[\sqrt{7-x}-2]=\frac{-1}{2\sqrt{7-x}}\)

derivative of denominator:

\(\displaystyle \L\\\frac{d}{dx}[1-\sqrt{4-x}]=\frac{1}{2\sqrt{4-x}}\)

\(\displaystyle \L\\\lim_{x\to\3}\frac{\frac{-1}{2\sqrt{7-x}}}{\frac{1}{2\sqrt{4-x}}}\)

=\(\displaystyle \L\\\lim_{x\to\3}\frac{-\sqrt{4-x}}{\sqrt{7-x}}\)

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