# We have #f:[0,2]->RR,f(x)=sqrt(4-x^2)#.How to solve this limit?#lim_(x->0)1/x^2int_0^xtf(t)dt#

The Final Limit has been evaluated using the following

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We see that:

Because we have this indeterminate form, we can use l'Hôpital's rule. Take the derivative of numerator and the denominator.

The derivative of the integral can be found through the Second Fundamental Theorem of Calculus.

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To solve the given limit, we can break it down into two parts: the limit as x approaches 0 of 1/x^2 and the integral part.

First, let's consider the limit as x approaches 0 of 1/x^2. Since x is approaching 0, we can see that 1/x^2 will tend towards positive infinity.

Next, let's focus on the integral part. We have the integral from 0 to x of tf(t) dt. To solve this, we can use the Fundamental Theorem of Calculus. By applying this theorem, we can differentiate the function inside the integral with respect to t and then evaluate it at the upper limit (x) and lower limit (0).

Differentiating f(t) = sqrt(4 - t^2) with respect to t, we get f'(t) = -t/sqrt(4 - t^2).

Now, we can evaluate the integral by substituting the limits into the derivative we obtained: ∫[0,x] tf(t) dt = [(-t^2)/2] from 0 to x = -(x^2)/2.

Combining the two parts, we have the limit as x approaches 0 of (1/x^2) * ∫[0,x] tf(t) dt = (1/x^2) * (-(x^2)/2) = -1/2.

Therefore, the solution to the given limit is -1/2.

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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.

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