How do you use continuity to evaluate the limit #arctan(x^2-4)/(3x^2-6x)#?

Question

Cameron Alexander · Accepted Answer

If #x->a#, where #a!=0# and #a!=2#, then #lim_{x->a}arctan(x^2-4)/(3x^2-6x)=arctan(a^2-4)/(3a^2-6a)# The function #f(x)=arctan(x^2-4)/(3x^2-6x)=arctan(x^2-4)/(3x(x-2))# is known to be continuous everywhere except #x=0# and #x=2# (this could be proved, but it would be a lot of work). To "evaluate a limit by using continuity" means that you can evaluate the limit of a known continuous function at a point of continuity just by substituting the point into the function. That is, if #f# is defined near #x=a# and continuous at #x=a#, then #lim_{x->a}f(x)=f(a)#. Applying this fact to the function #f# gives: #lim_{x->a}f(x)=lim_{x->a}arctan(x^2-4)/(3x^2-6x)=f(a)=arctan(a^2-4)/(3a^2-6a)# when #a!=0# and #a!=2#. This is not part of the given question, but it turns out that, even though #f(2)# does not exist and #f# is not continuous at #x=2#, the limit #lim_{x->2}f(x)# does exist in this example. In fact, from L'Hopital's Rule , it can be shown that #lim_{x->2}arctan(x^2-4)/(3x^2-6x)=2/3#. Try seeing if you can prove this!

Ezra Adams · Answer

undefined To use continuity to evaluate the limit of arctan(x^2-4)/(3x^2-6x), we can first simplify the expression by factoring out a common factor of x from both the numerator and denominator. This gives us arctan(x^2-4)/(3x(x-2)).

Next, we can evaluate the limit by plugging in the value that x approaches into the simplified expression. In this case, we are interested in finding the limit as x approaches a certain value, let's say c.

To evaluate the limit, we can use the fact that the arctan function is continuous for all real numbers. This means that we can evaluate the limit by directly substituting the value c into the expression. So, we substitute x = c into the simplified expression:

arctan(c^2-4)/(3c(c-2))

By doing this, we obtain the value of the limit as x approaches c.