# How do you evaluate the limit #root4(2cosx-5)# as x approaches 0?

The limit does not exist in

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To evaluate the limit of √(4(2cosx-5)) as x approaches 0, we substitute the value of x into the expression.

√(4(2cos(0)-5)) simplifies to √(4(2-5)) which further simplifies to √(4(-3)).

Since the square root of a negative number is not defined in the real number system, the limit does not exist.

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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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- For what values of x, if any, does #f(x) = 1/((x+8)(x-7)) # have vertical asymptotes?
- What is the limit of #(sqrt x) / (x + 4)# as x approaches infinity?
- How do you evaluate the limit #(2sinx-sin2x)/(xcosx)# as x approaches #0#?

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