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Chapter 3: Problem 41

Find the horizontal asymptote, if there is one, of the graph of each rational function. $$ h(x)=\frac{12 x^{3}}{3 x^{2}+1} $$

Short Answer

Expert verified
Based on the solution above, there are no horizontal asymptotes for this function because as x approaches both positive and negative infinity, the function also approaches positive and negative infinity, respectively.

Step by step solution

01

Simplify the function

First, we can simplify the function by factoring out the highest power of x in the numerator and the denominator, which in this case is \(x^{2}\). So, the function becomes: \(h(x) = \frac{{12x(x^{2})}}{{x^{2}(3 + \frac{1}{{x^{2}}})}} = \frac{{12x}}{{3 + \frac{1}{{x^{2}}}}}\).
02

Find limit as x approaches positive infinity

We need to determine the limit of the function as x approaches positive infinity. This is done through observing what happens to the function as x gets larger and larger. For this function, as x approaches infinity, the term \(\frac{1}{{x^{2}}}\) approaches 0 since any number divided by infinity is effectively zero. Therefore the function simplifies to: \(\frac{{12x}}{3}\) or \(4x\). So as x approaches positive infinity, the function approaches \(4 \cdot \infty\) which is also infinity.
03

Find limit as x approaches negative infinity

Similarly, we need to determine the limit of the function as x approaches negative infinity. This is the same process as step 2, but instead we are observing what happens to the function as x gets smaller and smaller. Just like in step 2, as x approaches negative infinity, the term \(\frac{1}{{x^{2}}}\) still approaches 0. So, the function simplifies to \(\frac{{12x}}{3}\) or \(4x\). However, as x approaches negative infinity, the functions approaches \(4 \cdot -\infty\) which is negative infinity.

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