Evaluate Limit: x² - 10x + 25/x-5

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In summary, the conversation discusses the evaluation of a limit and the use of absolute value when simplifying the expression. It is determined that the limit as x approaches 5 from the right is 1, the limit as x approaches 5 from the left is -1, and the overall limit does not exist due to the limits from the left and right not being equal.
  • #1
davemoosehead
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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?
 
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  • #2
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.
 
  • #3
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?
 
  • #4
The square root of a number is defined to be the positive square root. The numerator is |x-5|.
 
  • #5
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?
 
  • #6
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.
 
  • #7
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.
 
  • #8
Bingo, you've got it.
 
  • #9
Thank you so much
 

FAQ: Evaluate Limit: x² - 10x + 25/x-5

What is the limit of the function as x approaches 5?

The limit of the function f(x) = x² - 10x + 25 / x-5 as x approaches 5 is undefined. This is because when x approaches 5, the denominator becomes 0, which is an undefined value in mathematics.

How can I evaluate the limit of the function?

To evaluate the limit of the function f(x) = x² - 10x + 25 / x-5, you can use the concept of factorization. By factoring the numerator, you can simplify the function to f(x) = (x-5)(x-5) / x-5. Then, you can cancel out the common factors of x-5 to get f(x) = x-5. Finally, you can substitute the value of 5 for x to get the limit as 0.

Can the limit of this function be evaluated using direct substitution?

No, the limit of the function f(x) = x² - 10x + 25 / x-5 cannot be evaluated using direct substitution. As mentioned earlier, this is because when x approaches 5, the denominator becomes 0, which is an undefined value. Direct substitution can only be used when the function is defined and continuous at the given value.

What is the significance of the function being undefined at x=5?

The function being undefined at x=5 means that the function is not continuous at this point. This can indicate a potential break or discontinuity in the graph of the function, which may affect the overall behavior of the function.

Is it possible to manipulate the function to make the limit defined at x=5?

Yes, it is possible to manipulate the function to make the limit defined at x=5. In this case, you can use the concept of a removable discontinuity. By simplifying the function and canceling out the common factors of x-5, you can create a new function that is defined and continuous at x=5. However, this new function may have a different value compared to the original function at other points, so it is important to consider the overall behavior of the function before manipulating it.

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