find the limit or show that it doesn't exist

Question

lim x--> infinity (3x-2)/(2x+1)

 

so first i divided everything by x and then got.....

[3-(2/x)/2+(1/x)]

 

i didn't know where to take it from there or if i am even taking the correct primary steps 

Answer 1

Notice that taking the limit as x approaches infinity is when you find the horizontal asymptote.  In precalculus, you would divide the coefficients of the leading term if the degree of both the numerator and denominator are the same, in order to find the horizontal asymptote.

 

Therefore,

 

horizontal asymptote = 3/2 = limit as x approaches infinity

Answer 2

Since this is an indeterminate form, you can also use L'Hopital's rule.  Have you covered it in your calculus class?  The rule will give a limit of 3/2 as x→∞.

Answer 3

You did complete the primary steps correctly, Julia. Now, ask yourself what happens to 2/x and 1/x as x approaches infinity. The numerator of each of those terms is constant. However, the denominator of each is growing without bound. That means that both 2/x and 1/x approach 0 as x approaches infinity. IF the denominator of a fraction gets larger and larger, and the numerator is constant, then the overall value of the fraction will get smaller and smaller. 

 

Therefore, the limit of your function is just 3/2. You could graph this rational function and examine its behavior. You should be able to see that as x goes to infinity, the graph will approach 3/2. It will also have a vertical asymptote at x = -1/2. That's because x = -1/2 is not in the domain of this function. Why? Because when x = -1/2, then the denominator will equal 0, and division by 0 is undefined.