Another Limit Algebraically with SQRT in ti

[hopelynnwelch]

I have had no problems doing these until there are SQRTs involved.

The limit as x goes to 6 from the negative side of (sqrt(6-x))/(6-x)

I'm trying this again and if I multiply this function by (sqrt(6-x))/(sqrt(6-x)) and don't multiply out the denominator and then cancel I get

1/(sqrt(6-x))

This is undefined at 6 because 1/0 is undefined...

What did I miss here?

I should edit this to add that I know the limit is INFINITY. But how the heck can I come to this conclusion algebraically?

 

[Ishuda]

I have had no problems doing these until there are SQRTs involved.

The limit as x goes to 6 from the negative side of (sqrt(6-x))/(6-x)

I'm trying this again and if I multiply this function by (sqrt(6-x))/(sqrt(6-x)) and don't multiply out the denominator and then cancel I get

1/(sqrt(6-x))

This is undefined at 6 because 1/0 is undefined...

What did I miss here?

I should edit this to add that I know the limit is INFINITY. But how the heck can I come to this conclusion algebraically?

You figured out the answer but just to present it another way: If we let t = \(\displaystyle \sqrt{6-x}\). then as \(\displaystyle x \to 6-\), i.e. x goes to 6 from the negative side, \(\displaystyle t \to 0+\) and
\(\displaystyle \frac{\sqrt{6-x}}{6 - x} = \frac{t}{t^2} = \frac{1}{t}\)
Thus
\(\displaystyle \lim_{x \to 6-} \frac{\sqrt{6-x}}{6 - x} = lim_{t \to 0+} \frac{1}{t} = +\infty\)

 

[pka]

I have had no problems doing these until there are SQRTs involved.

The limit as x goes to 6 from the negative side of (sqrt(6-x))/(6-x)

I'm trying this again and if I multiply this function by (sqrt(6-x))/(sqrt(6-x)) and don't multiply out the denominator and then cancel I get

1/(sqrt(6-x))

This is undefined at 6 because 1/0 is undefined...

What did I miss here?

As \(\displaystyle x\to 6^-\) then \(\displaystyle \sqrt{6-x}\to 0^+\).
So the limit is \(\displaystyle +\infty\).

 

[Steven G]

I have had no problems doing these until there are SQRTs involved.

The limit as x goes to 6 from the negative side of (sqrt(6-x))/(6-x)

I'm trying this again and if I multiply this function by (sqrt(6-x))/(sqrt(6-x)) and don't multiply out the denominator and then cancel I get

1/(sqrt(6-x))

This is undefined at 6 because 1/0 is undefined...

What did I miss here?

I should edit this to add that I know the limit is INFINITY. But how the heck can I come to this conclusion algebraically?

you should know that sqrt(x-3)/(x-3) = 1/sqrt(x-3). No need to multiply top and bottom by anything. A major prerequisite for calculus 1 is algebra. If you do not know your algebra perfectly then maybe you should review your algebra.

1/0+ = infinity