Limits Involving Infinity. Example Problem 2
In this case, we’re dealing with a negative infinity. The graph could have no asvmptotes, or a horizontal asymptote of zero or some finite value of y. We will now look at the largest degree of the numerator and denominator. The largest degree of the numerator is four. The denominator, though, can be tricky because it includes a negative infinity. This makes sense because the graph of e to the x looks like this: lim_(x->infinity)(x^4-3x^2-Sx)/(7x^4+9x^3+e^x) [Graph] Behaves like x^4/7x^4 -> 1/7 As x gets infinitely large, as we increase the outbound, the y value is going to increase without bound. However, as we move to the left, as x approaches negative infinity, the y value does not decrease without bound. Instead you can see the behavior of a horizontal asymptote. The y value tends to zero. So, if we approached infinity in this problem, the denominator degree would be larger than the numerator degree, making this a bottom-heavy function and thus approaching a horizontal asymptote of zero. But when e to the x is extremely small, it behaves like zero, effectively making the denominator equal to 4. If you look at the behavior of the function and consider its power to the fourth over seven to the fourth, then this reduces to one seventh. If you think of the graph of one seventh, you see 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 even though the task did not ask this, if x were moved to the right of infinity, the limit would still be one 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 to the fourth is equal to one over seven. to The limit 1S one seventh.
