Limits and difference of squares help

[LutherBaker]

Homework Statement

find the limit as x tends to 3 of [sqrt(2x+3)-x] / (x-3)

Homework Equations

The Attempt at a Solution

This is from an old Protter textbook I am working through. I started with the difference of squares which results in

[2x + 3 - x^2]/ [(x-3)*sqrt(2x+3)+x]

but now I'm stuck. I can't seem to factor (x-3) out of the numerator.

 

[Dick]

Homework Statement

find the limit as x tends to 3 of [sqrt(2x+3)-x] / (x-3)

Homework Equations

The Attempt at a Solution

This is from an old Protter textbook I am working through. I started with the difference of squares which results in

[2x + 3 - x^2]/ [(x-3)*sqrt(2x+3)+x]

but now I'm stuck. I can't seem to factor (x-3) out of the numerator.

Well, you can. 2x+3-x^2 is equal to zero if you put x=3. Hence it must have a factor of x-3.

 

[Mark44]

Homework Statement

find the limit as x tends to 3 of [sqrt(2x+3)-x] / (x-3)

Homework Equations

The Attempt at a Solution

This is from an old Protter textbook I am working through. I started with the difference of squares which results in

[2x + 3 - x^2]/ [(x-3)*sqrt(2x+3)+x]

but now I'm stuck. I can't seem to factor (x-3) out of the numerator.

I don't see any ways that involve the conjugate or factoring, but L'Hopital's Rule gives a result.

 

[LutherBaker]

Well, you can. 2x+3-x^2 is equal to zero if you put x=3. Hence it must have a factor of x-3.

Got it! That was all I needed.

Turns out that 2x+3-x^2 can be refactored a few ways. First of all, I rearranged the components to -x^2+2x+3 and then I determined both (-x+3)(x+1) ... and now I see it ... (x-3)(-x-1). That allows me to solve the limit correctly. I just didn't try hard enough with the -x^2 in there.

Regarding your point above here, does this property have a name? IE: if you can replace x with a number, then by definition, the function can be refactored into (x-#)(...). I've not noticed/appreciated that property before.

 

[LutherBaker]

I don't see any ways that involve the conjugate or factoring, but L'Hopital's Rule gives a result.

Someone else mentioned that here at work but was quick to ask if the text had reviewed it. I'm only on chapter 2 so I think that L'Hopital's approach might avoid what the text, at this point, is trying to convey. Namely, learning how to refactor to find limits.

I'm just guessing though - its an old book. Thanks for the input!

 

[Mark44]

I like Dick's approach better - I prefer a factoring approach over using L'Hopital's Rule, since factoring is in a sense, a simpler approach.

BTW, in my first college calculus class we used one edition of the Protter & Morrey book. I still have it. It's very different from contemporary calculus texts - few illustrations, none in color. I probably paid less than $10 for it new.