How do you find the limit of # (sqrt(3x - 2) - sqrt(x + 2))/(x-2)# as x approaches 2?

Question

How do you find the limit of  # (sqrt(3x - 2) - sqrt(x + 2))/(x-2)# as x approaches 2?

Isabella Ackerman · Accepted Answer

#lim_(x->2)(sqrt(3x-2)-sqrt(x+2))/(x-2)=1/2#

The first thing we do in limit problems is substitute the #x# value in question and see what happens. Using #x=2#, we find: #lim_(x->2)(sqrt(3x-2)-sqrt(x+2))/(x-2)=(sqrt(3(2)-2)-sqrt((2)+2))/((2)-2)=(sqrt(4)-sqrt(4))/0=0/0#

You may be wondering how this helps us. Well, because we have #0/0#, this problem becomes fair game for an application of L'Hopital's Rule. This rule says that if we evaluate a limit and get #0/0# or #oo/oo#, we can find the derivative of the numerator and denominator and try evaluating it then.

So, without further ado, let's get to it.

Derivative of Numerator We're trying to find #d/dx (sqrt(3x-2)-sqrt(x+2))# here. Using the sum rule, we can simplify this to #d/dxsqrt(3x-2)-d/dxsqrt(x+2)#. Taking the derivative of the first term, we see: #d/dxsqrt(3x-2)=(3x-2)^(1/2)=3/(2sqrt(3x-2))->#Using power rule and chain rule

For the second term, #d/dxsqrt(x+2)=(x+2)^(1/2)=1/(2sqrt(x+2))->#Using power rule

Thus, our new numerator is: #3/(2sqrt(3x-2))-1/(2sqrt(x+2))#.

Derivative of Denominator This one is fairly easy: #d/dx(x-2)=1#. Yep, that's it.

Put it all Together Combining these two results into one, our new fraction is #(3/(2sqrt(3x-2))-1/(2sqrt(x+2)))/1=3/(2sqrt(3x-2))-1/(2sqrt(x+2))#. We can now evaluate it at #x=2# and see if anything changes: #lim_(x->2)(3/(2sqrt(3x-2))-1/(2sqrt(x+2)))=3/(2sqrt(3(2)-2))-1/(2sqrt((2)+2))=3/(2sqrt(4))-1/(2sqrt(4))=1/2#

And there you have it! We can confirm this result by looking at the graph of #(sqrt(3x-2)-sqrt(x+2))/(x-2)#: graph{(sqrt(3x-2)-sqrt(x+2))/(x-2) [-0.034, 3.385, -0.205, 1.504]}

Scarlett Allison · Answer

If your calculus course has not yet covered derivatives and l'Hospital's rule, use algebra to rationalize the numerator.

#(sqrt(3x-2)-sqrt(x+2))/(x-2) = (sqrt(3x-2)-sqrt(x+2))/(x-2) *(sqrt(3x-2)+sqrt(x+2))/(sqrt(3x-2)+sqrt(x+2))# # = ((3x-2)-(x+2))/((x-2)(sqrt(3x-2)+sqrt(x+2)))# # = (2x-4)/((x-2)(sqrt(3x-2)+sqrt(x+2)))# # = (2cancel((x-2)))/(cancel((x-2))(sqrt(3x-2)+sqrt(x+2)))# So we have #lim_(xrarr2)(sqrt(3x-2)-sqrt(x+2))/(x-2) = lim_(xrarr2)2/(sqrt(3x-2)+sqrt(x+2))# # = 2/(sqrt(3(2)-2)+sqrt((2)+2))# # = 2/(sqrt(4)+sqrt(4)) = 2/(2+2) = 1/2#

Axel Alvarez · Answer

undefined To find the limit of the given expression as x approaches 2, we can simplify it using algebraic manipulation. By multiplying the numerator and denominator by the conjugate of the numerator, we can eliminate the square roots.

The conjugate of the numerator is (sqrt(3x - 2) + sqrt(x + 2)).

Multiplying the numerator and denominator by the conjugate, we get:

[(sqrt(3x - 2) - sqrt(x + 2))/(x-2)] * [(sqrt(3x - 2) + sqrt(x + 2))/(sqrt(3x - 2) + sqrt(x + 2))]

Expanding the numerator using the difference of squares, we have:

[(3x - 2) - (x + 2)] / [(x - 2)(sqrt(3x - 2) + sqrt(x + 2))]

Simplifying further, we get:

(2x - 4) / [(x - 2)(sqrt(3x - 2) + sqrt(x + 2))]

Now, we can cancel out the common factor of (x - 2) in the numerator and denominator:

(2(x - 2)) / [(x - 2)(sqrt(3x - 2) + sqrt(x + 2))]

Simplifying again, we have:

2 / (sqrt(3x - 2) + sqrt(x + 2))

Now, we can substitute x = 2 into the expression:

2 / (sqrt(3(2) - 2) + sqrt(2 + 2))

2 / (sqrt(4) + sqrt(4))

2 / (2 + 2)

2 / 4

1/2

Therefore, the limit of the given expression as x approaches 2 is 1/2.