L'Hopital's Rule case: How does x^(-4/3) equal 0 when x approches infinity?

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  • #1
Mohmmad Maaitah
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Homework Statement
How is this 0/0 so we can use L'Hopital's Rule?
Relevant Equations
L'Hopital's Rule
I'm talking about the x^(-4/3) how does it equal 0 when x approch infinite??
so I can use

L'Hopital's Rule

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  • #2
$$x^{\frac{4}{3}}\rightarrow +\infty$$
so its inverse
$$x^{-\frac{4}{3}}\rightarrow +0$$
 
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FAQ: L'Hopital's Rule case: How does x^(-4/3) equal 0 when x approches infinity?

What is L'Hopital's Rule?

L'Hopital's Rule is a mathematical method used to evaluate limits of indeterminate forms, such as 0/0 or ∞/∞. It states that if the limits of functions f(x) and g(x) as x approaches a value c are both 0 or both ∞, then the limit of f(x)/g(x) as x approaches c can be found by taking the limit of their derivatives, provided this limit exists.

Why is x^(-4/3) considered in the context of L'Hopital's Rule?

In the context of L'Hopital's Rule, x^(-4/3) might be considered when evaluating limits that result in indeterminate forms. Understanding the behavior of x^(-4/3) as x approaches infinity helps determine if L'Hopital's Rule can be applied or if another method should be used to evaluate the limit.

How does x^(-4/3) behave as x approaches infinity?

As x approaches infinity, x^(-4/3) approaches 0. This is because the exponent -4/3 is negative, indicating that the function is a decreasing power function. As x becomes very large, the value of x^(-4/3) becomes very small, tending towards 0.

Why does x^(-4/3) approach 0 and not infinity when x approaches infinity?

x^(-4/3) approaches 0 because the exponent -4/3 is negative. For any positive number x, raising it to a negative power results in a fraction (1 over the positive power). As x grows larger, this fraction becomes smaller and smaller, approaching 0.

Can L'Hopital's Rule be applied directly to limits involving x^(-4/3)?

L'Hopital's Rule can be applied to limits involving x^(-4/3) if the limit results in an indeterminate form such as 0/0 or ∞/∞. However, it's essential to first confirm that the limit indeed results in an indeterminate form and that the derivatives of the numerator and denominator exist and are continuous around the point of interest.

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