Evaluate

Question

Evaluate

\lim_{x \to \infty} f (x)

and

\lim_{x \to - \infty} f (x)

.

$f (x) = \frac{4 x^{3} + 1}{1 - x^{3}}$

Accepted Answer

Welcome back, everyone. Evaluate the limits of FX as X approaches positive and negative infinity for the function F of X equals 5 XQ plus 1 divided by XQ to minus 1. So let's evaluate the first limit that says Li SX approaches positive infinity of the function, and our function is 5 x cubed plus 1 divided by X cubed minus 1. This is a rational function. What we're going to do is apply the right substitution, which gives us an indeterminate form, infinity divided by infinity. Whenever we get this form and we have a rational function, what we're going to do is basically notice that the numerator and the denominator, well, those are polynomials. So we want to identify the highest exponent. Of X and that's X cubed, right, and this means that we want to divide all of our terms by X cubed. This will allow us to deal with the indeterminate form and to evaluate the limit. So let's go ahead and do that. We're going to write 5X cubed, which is divided by X cubed, then we are dividing 1 by X cubed, and we have to perform the same steps for the denominator. X cubed divided by X cubed minus 1 divided by X cubed. Let's simplify to clearly see the final expression. Limit as x approaches infinity of 5 plus 1 divided by x cubed. That's our numerator, and the denominator is 1 minus 1 divided by X cubed. As X approaches infinity, 1 divided by X cubed approach is 0, right? It doesn't matter if we're going from the left or from the right because in front we have whole numbers that are added to each term. And this means that the limit is equal to 5. That's because we have a limit of 5 divided by 1. Well, we're adding 0 and then we are subtracting 0. But overall, That's 5 divided by 1, which is a finite number. And this means that the first limit is equal to 5. For the second limit, we're going to use the simplified form limit as X approach is negative infinity, and our simplified form is 5 + 1 divided by X cubed. Divided by 1 minus. 1 divided by x cut and in this case x approaches negative infinity so now. Nothing really changes because we have 5 + 1 divided by negative infinity, and that's basically 0. That's how we can visualize that. And for the denominator, we have 1 minus 1 divided by negative infinity, right? So that's negative. 0, right, because we are approaching 0 from the left. But once again we notice that this has no effect. It doesn't really matter if we are approaching 0 from the left or from the right because we have two whole numbers in front, and this means that once again we have 5 divided by 1, which gives us 5. So the answer to this problem is 5 for each limit. Limit as X approaches positive or negative infinity of F of X is equal to 5. Thank you for watching.

Evaluate limx→∞f(x){\displaystyle\lim_{x\to\infty}{f(x)}} andlimx→−∞f(x){\displaystyle\lim_{x\to-\infty}{f(x)}}.f(x)=4x3+11−x3f\left(x\right)=\frac{4x^3+1}{1-x^3}

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Evaluate $\lim_{x \to \infty} f (x)$ and $\lim_{x \to - \infty} f (x)$ .

$f (x) = \frac{4 x^{3} + 1}{1 - x^{3}}$