Limits Involving Infinity (Horizontal and Vertical Asymptotes Revisited) Limits as Approaches Infinity At times you ll need to know the behavior of a function or an epression as the inputs get increasingly larger larger in the positive and negative directions We can evaluate this using the it f( ) and f( ) Obviously, you cannot use direct substitution when it comes to these its Infinity is not a number, but a way of denoting how the inputs for a function can grow without any bound You see its for approaching infinity used a lot with fractional functions E) Evaluate 1 using a graph A more general version of this it which will help us out in the long run is this GENERALIZATION For any epression (or function) in the form CONSTANT = n CONSTANT, this it is always true POWER OF X
HOW TO EVALUATE A LIMIT AT INFINITY FOR A RATIONAL FUNCTION Step 1: Take the highest power of in the function s denominator and divide each term of the fraction by this power Step 2: Apply the it to each term in both numerator and denominator and remember: n C / = 0 and C = C where C is a constant Step 3: Carefully analyze the results to see if the answer is either a finite number or or E) Evaluate the it 6 3 5 + 2 E) Evaluate the it 3 2 5 2 + 7 2
E) Evaluate the it 2 5 + 2 3 2 2 E) Evaluate the it + 4 2 2 5 3 3 4 + 2 7
Limits at Vertical Asymptotes Now let s go back to its where approaches a specific number instead of ± You ll always want to try direct substitution first We ve seen what happens when the direct substitution leads to 0/0 there s usually an algebraic or trigonometric remedy to fi the it non-zero constant However, if your direct substitution leads to the form zero then you re trying to take a it where the function has a vertical asymptote Your answer to a it is still going to represent the y-value or y-direction the function takes as approaches a so in the event of a vertical asymptote the answer will usually involve ± E) Evaluate WITHOUT using a graph (analyze the its logically) a) b) 5 5 5+ 5 c) What conclusion can be made for 5 5?
Now that you ve seen the semantics behind evaluating these its, you ll have a better understanding about what the function s graph will look like The its we just evaluated indicate the vertical asymptote behavior both left and right of = 5 If we also looked at the it as (right end behavior) and the it as (left end behavior) we would have determined that the function has a horizontal asymptote at y = 1 / 1 1 = = = = 1 5 / 5/ 1 5/ 1 0 HA at y = 1 So, going back to our original it strategy, if you try a direct substitution and you get a fraction in the form NON-ZERO CONSTANT / ZERO, then you re approach is located at one of the function s vertical asymptotes and your answer will be either or STEPS FOR EVALUATING LIMITS AT VERTICAL ASYMPTOTES Step 1: Make sure the fractional epression in your it is completely factored Step 2: Use direct substitution on all the factors which DO NOT BECOME ZERO make special note of the signs of these numbers you get Step 3: One of the denominator s factors will evaluate as 0 so we ll need to analyze it more carefully Based on which one-sided it (left or right approach) you re evaluating, this factor approaching zero will be will be positive or negative and VERY SMALL This VERY SMALL value in the denominator will cause the evaluation of the it to go infinite Step 4: Considering all the factors your evaluated in step 2 and the factor approaching 0 you analyzed in step 3: if there are an even number of negative factors the it will evaluate as if there are an odd number of negative factors the it will evaluate as
E) Evaluate the its 2 + 1 a) 4 3 ( + 3) 2 + 1 b) + 4 3 ( + 3) 2 + 1 c) Make a conclusion for 3 ( 3) 4 +
E) Evaluate 6 3 + 2 4 2 9 2 for + a) 3 b) 3 c) 3 + d) 0 e) 0 f) 0