Video: Finding the Limit at Infinity of a Combination of Rational Functions

Find lim_(π‘₯ β†’ ∞) (βˆ’4/π‘₯Β² + 5/π‘₯ + 8).

02:17

Video Transcript

Find the limit of negative four over π‘₯ squared plus five over π‘₯ plus eight as π‘₯ approaches infinity.

We have a limit as π‘₯ approaches infinity here, but all the normal rules of limits still apply. For example, the limit of a sum of functions is equal to the sum of the limits. And so, we can split our limit up into three. It’s equal to the limit of negative four over π‘₯ squared as π‘₯ approaches infinity plus the limit of five over π‘₯ as π‘₯ approaches infinity plus the limit of eight as π‘₯ approaches infinity.

What can we say about this limit? Well, we know that the limit of a constant 𝐾, as π‘₯ approaches some number 𝑐, is just 𝐾. And as for the previous limit law, this holds true, even if 𝑐 isn’t a real number but is infinity or negative infinity. The value of this last limit is just eight.

What about the other two limits? We can use the fact that the limit of a constant multiple of a function is that constant multiple of the limit of the function. The first limit is, therefore, negative four times the limit of one over π‘₯ squared as π‘₯ approaches infinity. And the second is five times the limit of one over π‘₯ as π‘₯ approaches infinity. And finally, we add the eight.

Now, the limit of the reciprocal function one over π‘₯, as π‘₯ approaches infinity, is something we should know. Its value is zero. But how about the limit of one over π‘₯ squared as π‘₯ approaches infinity? Well, we can use the fact that the limit of a power of a function is that power of the limit of the function. This limit is the limit of the reciprocal function one over π‘₯ squared, as one over π‘₯ squared equals one over π‘₯ all squared. And by our limit law, this is the limit of one over π‘₯ as π‘₯ approaches infinity all squared. This limit is known to be zero. And so, our limit, the limit of one over π‘₯ squared as π‘₯ approaches infinity, is also zero.

We can generalize in that you get another limit law that the limit of one over π‘₯ to the power of 𝑛, as π‘₯ approaches infinity, is zero, at least if 𝑛 is greater than zero. Our original limit is, therefore, negative four times zero plus five times zero plus eight, which is, of course, just eight.

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