Finding Limits Analytically. Example Problem 2

Finding Limits Analytically. Example Problem 2

Find:
lim_(x->3) (sqrt(2x+10-4))/x-3 * (sqrt(2x+10+4))/(sqrt(2x+10+4))
lim_(x->3) (2x+10-16)/((x-3)(sqrt(2x+10+4)))
lim_(x->3) 2x-6/(x-3)(sqrt(2x+10+4))
lim_(x->3) 2/(sqrt(2x+10+4))=2/8=1/4
Another algebraic technique we can use to clear up limits of indeterminate forms, 0/0, is to
rationalize the denominator.
Clearly, factoring the denominator is not going to work here. The square root of a negative
number is also a negative number and therefore cannot be factored.

Since we cannot factor
the denominator, we must rationalize it. To rationalize a fraction, we multiply both the
numerator and denominator by their conjugate and then simplify the answer.
To calculate the value of the expression, we multiply the following values: the square root of
2 times x squared plus 10 times x plus 4 (the first term) and then the square root of 2 times x
squared plus 10 times x plus 4.
We have a difference of squares, which can be solved by adding or subtracting a perfect
square from both sides.

So the numerator is 2x plus 10 minus 16. Don't multiply this
together at all. x minus 3 and square root 2x plus 10 plus 4.
The next step is to clean the numerator. 10 minus 16 is 6 over × minus 3 and square root 2x
plus 10 plus 4.
You'll notice 2x minus 6 and x minus 3 divide and have a remainder of 2, which is nice.
3 plugged into the numerator, which gave the result of 0. We've now taken out the 0/0.
Now that we have evaluated the denominator, we can take the limit again.

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Evaluating the denominator of a fraction, we multiply 2 by 3 to get 6, plus 10 to get 16. The
square root of 16 is 4, plus 4 plus 4 equals 8; therefore, the limit as x approaches 3 is 1/4.