Evaluate Limit: x² - 10x + 25/x-5

[davemoosehead]

Homework Statement

Evaluate the following limit, if possible:

lim sq(x² - 10x + 25)
x->5+ x - 5

Homework Equations

The Attempt at a Solution

(x-5)(x-5)
(x-5)

lim x - 5
x->5+

0?

x approaches 0? How does the + (one sided) come into play?

 

[Dick]

I assume the 'sq' means square root. If so can you express the numerator in terms of an absolute value? You'll find the value is quite different if x is a little less than five versus a little larger than five.

 

[davemoosehead]

yea i forgot about the square root

sqr[ (x-5)² ]
(x-5)

So the numerator would just be (x-5) after the sq is canceled out.

Are you saying since x approaches 5 from the right (positive), we only consider the absolute value of x-5? Why would I only apply the absolute value to the numerator?

 

[Dick]

The square root of a number is defined to be the positive square root. The numerator is |x-5|.

 

[davemoosehead]

Ok, that makes sense.

ok so it's reduced down to

|x-5|
x-5

i like to make tables:
x|y
3|-1
4|-1
5|0?
6|1
7|1

I remember my teacher saying something special about 0/0. Does that mean this limit does not exist?

So the limit as x->5+ does not exist.

Would it be correct to say the limit as x->5- does not exist?

 

[Dick]

Noooo! The limit as x->5+ is the limit of a sequence with x values like 5.1, 5.01, 5.001... The value of the function at the limit doesn't have to be defined for a limit to exist.

 

[davemoosehead]

Haha I think I get it now. When we say x->5+ we pick a value to the right of 5 and make it smaller so it approaches 5. So as x approaches 5, the limit is 1.

The limit of x->5- would be -1

And the limit of x->5 would be DNE because the limit from the left does not equal the limit from the right.

 

[Dick]

Bingo, you've got it.

 

[davemoosehead]

Thank you so much