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

Question

Evelyn Agard · Accepted Answer

A function #f(x)# is said to be continuous at #a# if #lim_(xrarra)f(x)=f(a)# So in this case, our #f(x)# is #sqrt(2x-3)# and #a# is #2#. Compute the limit, then verify that it equals #h(2)#: #color(white)=lim_(xrarr2)sqrt(2x-3)# #=sqrt(2(2)-3)# #=sqrt(4-3)# #=sqrt1# #=1# Here's #h(2)#: #h(2)=sqrt(2(2)-3)# #color(white)(h(2))=sqrt(4-3)# #color(white)(h(2))=sqrt1# #color(white)(h(2))=1# Since the limit and the function are equal, the function #h(x)# is continuous at #x=2#. Hope this helped!

Ezra Adams · Answer

undefined 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.