How do you find the limit of #(x^2-5x+4)/(x^2-2x-8)# as #x-4#?

Question

How do you find the limit of #(x^2-5x+4)/(x^2-2x-8)# as #x->4#?

Aurora Alvarado · Accepted Answer

Factor, simplify and try again.

When we try to evaluate by substitution, we get #lim_(xrarr4)(x^2-5x+4) = 0# and #lim_(xrarr4)(x^2-2x-8) = 0# . The initial form of #lim_(xrarr4)(x^2-5x+4) /(x^2-2x-8)# is the indeterminate form #0/0#. Because #4# is a zero of both polynomials (the numerator and the denominator), we can be sure that #x-4# is a factor of both. Therefore we can simplify the ration and try again to evaluate the limit. #lim_(xrarr4)(x^2-5x+4) /(x^2-2x-8) = lim_(xrarr4)((x-4)(x-1)) /((x-4)(x+2)) # # = lim_(xrarr4)(x-1) /(x+2) = (4-1)/(4-2) = 3/2 #

Axel Alvarez · Answer

undefined To find the limit of (x^2-5x+4)/(x^2-2x-8) as x approaches 4, we can substitute the value of 4 into the expression and simplify. By doing so, we get (4^2-5(4)+4)/(4^2-2(4)-8). Simplifying further, we have (16-20+4)/(16-8-8), which becomes 0/0. This is an indeterminate form. To evaluate the limit, we can factorize the numerator and denominator. Factoring the numerator gives us (x-4)(x-1), and factoring the denominator gives us (x-4)(x+2). Canceling out the common factor of (x-4), we are left with (x-1)/(x+2). Now, substituting the value of 4 into this simplified expression, we get (4-1)/(4+2), which equals 3/6 or 1/2. Therefore, the limit of (x^2-5x+4)/(x^2-2x-8) as x approaches 4 is 1/2.

How do you find the limit of #(x^2-5x+4)/(x^2-2x-8)# as #x-&gt;4#?

How do you find the limit of #(x^2-5x+4)/(x^2-2x-8)# as #x->4#?