How do you prove that the function #f(x) = [x^2 + x] / [x]# is not continuous at a =0?
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Expression undefined;
What I have in blue is indeterminate form. We have a zero in a denominator, which means this expression is undefined.
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To prove that the function f(x) = [x^2 + x] / [x] is not continuous at a = 0, we can show that the left-hand limit and the right-hand limit of the function at x = 0 are not equal.
First, let's find the left-hand limit (LHL) of the function as x approaches 0. We substitute values slightly less than 0 into the function and simplify:
lim(x→0-) [x^2 + x] / [x] = lim(x→0-) (x^2 + x) / x = lim(x→0-) (x(x + 1)) / x = lim(x→0-) (x + 1) = 1
Next, let's find the right-hand limit (RHL) of the function as x approaches 0. We substitute values slightly greater than 0 into the function and simplify:
lim(x→0+) [x^2 + x] / [x] = lim(x→0+) (x^2 + x) / x = lim(x→0+) (x(x + 1)) / x = lim(x→0+) (x + 1) = 1
Since the LHL and RHL are both equal to 1, we can conclude that the function f(x) = [x^2 + x] / [x] is continuous at a = 0.
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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.
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.
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