Would someone be able to explain why you are able to take the limit of each individual section of \( \lim _{x \rightarrow \infty} \frac{\frac{3}{x}-\frac{5}{x^{2}}}{1+\frac{7}{x^{3}}} \) to get zero? Like why we are able to know that \( \frac{3}{x},-\frac{5}{x^{2}}, 1 \), and \( \frac{7}{x^{3}} \) can be evaluated individually? Also, does this method only work when the equation is fully simplified?

To understand why we can take the limit of each individual term in the expression \[ \lim _{x \rightarrow \infty} \frac{\frac{3}{x}-\frac{5}{x^{2}}}{1+\frac{7}{x^{3}}} \] to get zero, we need to consider the behavior of each term as \( x \) approaches infinity. ### Step-by-Step Explanation 1. **Behavior of Individual Terms**: - As \( x \) approaches infinity, the term \( \frac{3}{x} \) approaches \( 0 \) because the numerator is constant (3) and the denominator (x) grows without bound. - Similarly, \( -\frac{5}{x^{2}} \) also approaches \( 0 \) as \( x \) increases, since \( x^{2} \) grows even faster than \( x \). - The constant term \( 1 \) remains \( 1 \) regardless of \( x \). - The term \( \frac{7}{x^{3}} \) approaches \( 0 \) as well, since \( x^{3} \) grows even faster than \( x \) or \( x^{2} \). 2. **Limit of the Entire Expression**: - Now, substituting these limits into the original expression, we have: \[ \lim _{x \rightarrow \infty} \frac{0 - 0}{1 + 0} = \frac{0}{1} = 0 \] ### Why This Works The reason we can evaluate the limit of each term individually is due to the properties of limits. Specifically, if the limit of the numerator and the limit of the denominator exist, we can evaluate the limit of the fraction as the limit of the numerator divided by the limit of the denominator: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{\lim_{x \to a} f(x)}{\lim_{x \to a} g(x)} \] provided that \( \lim_{x \to a} g(x) \neq 0 \). ### Simplification This method does not necessarily require the expression to be fully simplified, but it is essential that the limits of the individual terms exist. If the expression is complex or involves indeterminate forms (like \( \frac{0}{0} \)), then further simplification or different techniques (like L'Hôpital's Rule) may be necessary. ### Conclusion In summary, we can evaluate the limit of each term individually because they behave predictably as \( x \) approaches infinity. This method is valid as long as the limits of the individual terms exist and the denominator does not approach zero.