# How do you prove that #h(x) = sqrt(2x - 3)# is continuous as x=2?

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To prove that h(x) = sqrt(2x - 3) is continuous at x = 2, we need to show that the limit of h(x) as x approaches 2 exists and is equal to h(2).

First, let's find the value of h(2) by substituting x = 2 into the function: h(2) = sqrt(2(2) - 3) = sqrt(4 - 3) = sqrt(1) = 1

Next, we need to find the limit of h(x) as x approaches 2. We can do this by evaluating the limit expression: lim(x→2) sqrt(2x - 3)

To simplify this expression, we can substitute y = 2x - 3: lim(x→2) sqrt(y)

Now, as x approaches 2, y approaches 2(2) - 3 = 1. So, we can rewrite the limit expression as: lim(y→1) sqrt(y)

The square root function is continuous for all non-negative values of y. Since y = 1 is non-negative, we can conclude that: lim(y→1) sqrt(y) = sqrt(1) = 1

Since the limit of h(x) as x approaches 2 is equal to h(2), we can conclude that h(x) = sqrt(2x - 3) is continuous at x = 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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