# How do you determine the limit of #(x^2 -2x) / (x^2 - 4x + 4)# as x approaches 2-?

If we put in values close to 2 from the left of 2 like 1.9, 1.99..etc we see that our answer gets bigger in the negative direction going to negative infinity.

If you graph it as well you will see that as x comes to 2 from the left y drops without bound going to negative infinity.

You can also use L'Hopital's Rule but it will be the same answer.

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To determine the limit of (x^2 - 2x) / (x^2 - 4x + 4) as x approaches 2-, we substitute the value 2 into the expression. By plugging in 2 for x, we get (2^2 - 2(2)) / (2^2 - 4(2) + 4), which simplifies to (4 - 4) / (4 - 8 + 4). Further simplifying, we have 0 / 0. This is an indeterminate form. To evaluate the limit, we can factorize the denominator as (x - 2)^2. Canceling out the common factor of (x - 2), we are left with 1 / (x - 2). As x approaches 2-, the denominator approaches 0, and the limit becomes undefined.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- How do you find the limit of #(x^2-3x+2)/(x^3-4x)# as x approaches 2?
- How do you find the limit of #(2-x)/(x^2-4)# as #x->2#?
- How do you find the limit of #(sqrt(x) -sqrt 2)/ sqrt(x^2 - 2x)# as x approaches 2?
- How do you find the limit of #xtan(1/(x-1))# as x approaches infinity?
- How do you find the limit of #(x²-25)/(sqrt(2x+6)-4)# as x approaches 5?

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