Example Problem 2. Finding Limits Analytically

p {margin: 0; padding: 0;} .ft00{font-size:20px;font-family:Arial;color:#000000;} .ft01{font-size:16px;font-family:ArialMT;color:#000000;} .ft02{font-size:16px;line-height:21px;font-family:ArialMT;color:#000000;} Finding Limits Analytically. Example Problem 2 Another algebraic technique we can use to clear up limits of indeterminate forms, 0/0, is torationalize the denominator. Clearly, factoring the denominator is not going to work here.The square root of a negativenumber is also a negative number and therefore cannot be factored.Since we cannot factorthe denominator,we must rationalize it.To rationalize a fraction,we multiply both thenumerator 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 of2 times x squared plus 10 times x plus 4 (the first term) and then the square root of 2 times xsquared plus 10 times x plus 4.We have a difference of squares, which can be solved by adding or subtracting a perfectsquare from both sides. So the numerator is 2x plus 10 minus 16. Don't multiply thistogether 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 x minus 3 and square root 2xplus 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. Evaluating the denominator of a fraction, we multiply 2 by 3 to get 6, plus 10 to get 16. Thesquare root of 16 is 4, plus 4 plus 4 equals 8; therefore, the limit as x approaches 3 is 1/4.