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21. \( \lim _{w \rightarrow \infty} \frac{2 w^{2}-3 w+4}{5 w^{2}+7 w-1} \). 27. \( \lim _{x \rightarrow \infty} \frac{2 x^{2}}{(2 x-1)^{2}} \). \( \begin{array}{ll}\text { 25. } \lim _{x \rightarrow \infty} \frac{3}{\sqrt{x}} & \text { 24. } \lim _{x \rightarrow \infty} \frac{4-3 x^{3}}{x^{3}-1} \\ \text { 27. } \\ \text { 26. } \lim _{x \rightarrow \infty} \frac{(x-1)^{2}}{x^{3}+2} \\ \text { 2x } \\ \text { 2 }\end{array} \)

Ask by Romero Weaver. in China
Mar 17,2025

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Answer

The limits are: - **21.** \(\frac{2}{5}\) - **27.** \(\frac{1}{2}\) - **25.** \(0\) - **24.** \(-3\) - **26.** \(0\)

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The Deep Dive

When you're tackling limits involving polynomials or rational functions, it's often helpful to focus on the highest degree terms. For the first limit, both the numerator and the denominator are quadratic, so we only need to consider their leading coefficients. Simplifying gives you \( \lim_{w \to \infty} \frac{2w^2}{5w^2} = \frac{2}{5} \). In the limit \( \lim_{x \to \infty} \frac{2x^2}{(2x-1)^2} \), expanding the denominator leads to \( (2x-1)^2 = 4x^2 - 4x + 1 \). Focusing again on the leading term gives \( \lim_{x \to \infty} \frac{2x^2}{4x^2} = \frac{1}{2} \). Remember, in these types of limits, the highest degree terms dominate as \( x \) approaches infinity!

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