Problem 59 Hard Difficulty

A function $ f $ is a ratio of quadratic functions and has a vertical asymptote $ x = 4 $ and just one
$ x $-intercept, $ x = 1 $. It is known that $ f $ has a removable discontinuity at $ x = -1 $ and $ \displaystyle \lim_{x \to -1} f(x) = 2 $. Evaluate
(a) $ f (0) $ (b) $ \displaystyle \lim_{x \to \infty} f(x) $

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07:45

Daniel J.

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suppose we have a rational function with the following properties. The first one it has a quadratic numerator and denominator. Now, this means that the function can be written into a leading coefficient eight times we have x minus a Times X -7 over x minus c times x minus D. Next property is it has a vertical sm towed at x equals four. And this means that x minus four is a factor in the denominator and so we have F of X equal to, we have eight times x minus a times x minus B over Have X -4 times X -T. Next property is that X equals one. Is the only x intercept. Now having an X intercept. That means that this um X value the factor will be A factor in the numerator. So X -1. We will see this in the numerator. So you have F of X. This is equal to we have eight times x minus one times x minus b over x minus four times x minus d. Another property says that It has a removable discontinuity at x equals negative one. This means that the graph of the function has a hole at X equals -1. And if you have a whole, that means that there is a common factor in the numerator and denominator. That means Experts, one is a factor in both the numerator and denominator. And so our function now becomes F of X is equal to we have eight times X -1 times expose one Over. You have X -4 times X plus one. Now to find this a we will use the last property that says The limit of this function as X approaches -1 is equal to two. And so we have we have limits As X approaches -1 of this function eight times x -1 Times X -1 over X -4 times x plus one. This is equal to we have We can cancel out the express one and we get limit as X approaches negative one of eight times X -1 over X -4. Now this is equal to two and evaluating at -1, we get eight times negative 1 -1 Over negative 1 -4, That's equal to two. We get Negative to a over -5. This is equal to two and we get hey that's equal to five. And so the complete form of our function is F of X. This is equal to we have five times X -1 Times X-plus one over X -4 times x plus one. Or if we expanded to get five times x squared -1 over have x squared minus three X -4. Or this is the same as five, X squared minus five over x squared minus three x minus four. Now we want to find f of zero and so we have F of zero. This is five times zero squared minus five over. You have zero squared minus three times zero minus four. This is just negative five over negative four or 5/4. So this is the value of f of zero and the next one is to find the limit as X approaches infinity of f of X. And this is just limit as X approaches infinity of five, x squared -5 over X squared minus three X minus four. Now, if we are taking the limits at infinity we simply take her out the variable with the highest exponents for the numerator and denominator. So it will be limit as X approaches infinity of the factor at that square we get five minus five over x squared this all over you factor out excrete again for the denominator and we get 1 -3 over x minus four over X squared Now simplifying this. We get limit as X approaches infinity, this will reduce to 5 -5 over X squared over 1 -3 over X -4 over experts. Since the experts cancel out and evaluating at the infinity we get 5 -5 over infinity Over 1 -3 over infinity -4 over infinity and knowing that Constant over infinity approaches zero. So this becomes zero. This becomes zero and this also becomes zero. We get the limit of the function as X approaches infinity equal to 5/1 or that's five