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Limits Involving Infinity (Horizontal and Vertical Asymptotes Revisited)

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Limits Involving Infinity (Horizontal and Vertical Asymptotes Revisited) Limits Involving Infinity (Horizontal and Vertical Asymptotes Revisited) Limits as ‘ x ’ Approaches Infinity At times you’ll need to know the behavior of a function or an expression as the inputs get increasingly larger … larger in the positive and negative directions. We can evaluate this using the limit limf ( x ) and limf ( x ) . x→ ∞ x→ −∞ Obviously, you cannot use direct substitution when it comes to these limits. Infinity is not a number, but a way of denoting how the inputs for a function can grow without any bound. You see limits for x approaching infinity used a lot with fractional functions. 1 Ex) Evaluate lim using a graph. x→ ∞ x A more general version of this limit which will help us out in the long run is this … GENERALIZATION For any expression (or function) in the form CONSTANT , this limit is always true POWER OF X CONSTANT lim = x→ ∞ xn HOW TO EVALUATE A LIMIT AT INFINITY FOR A RATIONAL FUNCTION Step 1: Take the highest power of x in the function’s denominator and divide each term of the fraction by this x power. Step 2: Apply the limit to each term in both numerator and denominator and remember: n limC / x = 0 and lim C= C where ‘C’ is a constant. x→ ∞ x→ ∞ Step 3: Carefully analyze the results to see if the answer is either a finite number or ‘ ∞ ’ or ‘ − ∞ ’ 6x − 3 Ex) Evaluate the limit lim . x→ ∞ 5+ 2 x 3− 2x − 5 x 2 Ex) Evaluate the limit lim . x→ ∞ 2x + 7 5x+ 2 x −2 Ex) Evaluate the limit lim . x→ −∞ 3x2 − 2 x −2x4 − 5 x 2 + 3 x Ex) Evaluate the limit lim . x→ −∞ 4x3 + 2 x − 7 Limits at Vertical Asymptotes Now let’s go back to limits where x 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 ‘fix’ the limit. non-zero constant However, if your direct substitution leads to the form … zero … then you’re trying to take a limit where the function has a vertical asymptote . Your answer to a limit is still going to represent the y-value or y-direction the function takes as x approaches ‘a’ so in the event of a vertical asymptote the answer will usually involve ± ∞ . Ex) Evaluate WITHOUT using a graph (analyze the limits logically) x x a) lim b) lim x→5 − x − 5 x→5 + x − 5 x c) What conclusion can be made for lim ? x→5 x − 5 Now that you’ve seen the semantics behind evaluating these limits, you’ll have a better understanding about what the function’s graph will look like. The limits we just evaluated indicate the vertical asymptote behavior both left and right of x = 5. If we also looked at the limit as x → ∞ (right end behavior) and the limit as x → − ∞ (left end behavior) we would have determined that the function has a horizontal asymptote at y = 1. x x/x 1 1 lim= lim = lim == 1 H.A. at y = 1 x→∞x5− x →∞ x/x5/x − x →∞ 15/x −− 10 So, going back to our original limit strategy, if you try a direct substitution and you get a fraction in the form ‘NON-ZERO CONSTANT / ZERO’, then you’re x 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 expression in your limit 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 limit (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 limit 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 limit will evaluate as ‘ ∞ ’ … if there are an odd number of negative factors the limit will evaluate as ‘ − ∞ ’. Ex) Evaluate the limits 2x + 1 2x + 1 a) lim b) lim − 4 + 4 x→− 3 (x + 3) x→− 3 (x + 3) 2x + 1 c) Make a conclusion for lim . x→− 3 (x + 3) 4 6− 3x + 2 x 2 Ex) Evaluate lim for x4− 9 x 2 a) x → 3 + b) x → 3 − c) x → 3 d) x → 0 − e) x → 0 + f) x → 0 .
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