Example Problem 2. Limits Involving Infinity

p {margin: 0; padding: 0;} .ft00{font-size:20px;font-family:Arial;color:#000000;} .ft01{font-size:16px;font-family:ArialMT;color:#000000;} .ft02{font-size:16px;line-height:21px;font-family:ArialMT;color:#000000;} Limits Involving Infinity. Example Problem 2 In this case, we're dealing with a negative infinity. The graph could have no asymptotes, or a horizontal asymptote of zero or some finite valueof y.We will now look at the largest degree of the numerator and denominator.The largest degreeof the numerator is four.The denominator, though, can be tricky because it includes anegative infinity. This makes sense because the graph of e to the x looks like this: As x gets infinitely large, as we increase the outbound, the y value is going to increasewithout bound. However, as we move to the left, as x approaches negative infinity, the yvalue does not decrease without bound. Instead you can see the behavior of a horizontalasymptote. The y value tends to zero.So, if we approached infinity in this problem, the denominator degree would be larger thanthe numerator degree, making this a bottom-heavy function and thus approaching ahorizontal asymptote of zero.But when e to the x is extremely small, it behaves like zero, effectively making thedenominator equal to 4.If you look at the behavior of the function and consider its power to the fourth over seven tothe fourth, then this reduces to one seventh. If you think of the graph of one seventh, yousee that it is just a horizontal line.So, if we move x to the left of -infinity, we approach the y value of one seventh. And eventhough the task did not ask this, if x were moved to the right of infinity, the limit would still beone seventh. In this case, where the degree of the denominator and the numerator are exactly the same,we must take the ratio of the coefficients. We find that x to the fourth divided by seven x tothe fourth is equal to one over seven. The limit is one seventh.