# How do you evaluate the limit #lim(5x^2)/(2x)dx# as #x->0#?

The first line of attack when looking at a limit should be to plug in the number we want the function to approach.

This is an indeterminate form !

We solve indeterminate forms by using L'Hôpital's Rule.

If you're new to calculus, this might look a little messy, but I'll try to keep it simple, so bear with me. You can even skip the mumbo-jumbo and just read the math part if you don't need the theory.

then

Essentially, whichever one gets where it's going faster (i.e. has a greater rate of change) wins.

Now onto the math:

which means, by L'Hôpital's Rule:

note: There are also many cases in which you do L'Hôpital's Rule and get another indeterminate form, in which case you have to apply L'Hôpital's Rule more than once.

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To evaluate the limit lim(5x^2)/(2x)dx as x approaches 0, we can simplify the expression by canceling out the common factor of x in the numerator and denominator. This gives us the limit lim(5x)/(2).

Now, substituting x=0 into the expression, we get (5*0)/(2) = 0/2 = 0.

Therefore, the limit of (5x^2)/(2x)dx as x approaches 0 is equal to 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.

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