Limits

[PTstudent]

Hello I took a calc class a year ago and did not pass it. I am taking it again and want a good grade. Some of it is slowly coming back but I don't understand this question. lim x--> 9- (root(x))-3/ x-9 I'm having trouble with the square root sign. Would I just multiply the conjugate to the top and the bottom? That is the only thing I can think of. Any help would be appreciated.

 

[tkhunny]

\(\displaystyle \lim_{x\rightarrow 9}\frac{\sqrt{x}-3}{x-9}\)?

No, I do not think I would play with any "conjugate" of anything. My inclination would be to factor "x-9" as a Difference of Squares.

 

[burakaltr]

Hello I took a calc class a year ago and did not pass it. I am taking it again and want a good grade. Some of it is slowly coming back but I don't understand this question. lim x--> 9- (root(x))-3/ x-9 I'm having trouble with the square root sign. Would I just multiply the conjugate to the top and the bottom? That is the only thing I can think of. Any help would be appreciated.

sqrt(x)-3 / ( (sqrt(x)-3 )* (sqrt(x)+3) ) = (sqrtx + 3 )** -1 =


1/6 ---> answer

 

[lookagain]

sqrt(x)-3 / ( (sqrt(x)-3 )* (sqrt(x)+3) ) = (sqrtx + 3 )** -1 =


1/6 ---> answer

• There is a minus sign as a superscript. And you must use grouping symbols as:

lim x--> 9- [(root(x)-3]/(x-9)


\(\displaystyle \lim_{x\to\ 9^{-}} \ \frac{\sqrt{x} - 3}{x - 9}\)

 

[burakaltr]

• There is a minus sign as a superscript. And you must use grouping symbols as:

lim x--> 9- [(root(x)-3]/(x-9)


\(\displaystyle \lim_{x\to\ 9^{-}} \ \frac{\sqrt{x} - 3}{x - 9}\)

So, Sir

Do not you agree the answer is 1 / 6 ?

 

[lookagain]

So, Sir

Do not you agree the answer is 1 / 6 ?

I do agree, but leave out the spaces in your fractional answer
(or whenever you type fractions.)


Type this as 1/6, or as


\(\displaystyle 1/6,\) or as


\(\displaystyle \frac{1}{6}.\)