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

Question

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

UpStudy AI · Accepted Answer

**21.** Compute  
$$
\lim_{w\to\infty}\frac{2w^2-3w+4}{5w^2+7w-1}.
$$  
Since the highest power in the numerator and denominator is $$w^2$$, we divide each term by $$w^2$$:
$$
\frac{2w^2-3w+4}{5w^2+7w-1} = \frac{2-\frac{3}{w}+\frac{4}{w^2}}{5+\frac{7}{w}-\frac{1}{w^2}}.
$$
Taking the limit as $$w\to\infty$$ (so that $$\frac{3}{w},\, \frac{4}{w^2},\, \frac{7}{w},$$ and $$\frac{1}{w^2}$$ all go to 0):
$$
\lim_{w\to\infty}\frac{2w^2-3w+4}{5w^2+7w-1} = \frac{2}{5}.
$$

---

**27.** Compute  
$$
\lim_{x\to\infty}\frac{2x^2}{(2x-1)^2}.
$$
First, expand the denominator:
$$
(2x-1)^2 = 4x^2 - 4x + 1.
$$
Now, divide the numerator and denominator by $$x^2$$:
$$
\frac{2x^2}{4x^2-4x+1} = \frac{2}{4-\frac{4}{x}+\frac{1}{x^2}}.
$$
Taking the limit as $$x\to\infty$$ (observing that $$\frac{4}{x}\to 0$$ and $$\frac{1}{x^2}\to 0$$):
$$
\lim_{x\to\infty}\frac{2x^2}{(2x-1)^2} = \frac{2}{4} = \frac{1}{2}.
$$

---

**25.** Compute  
$$
\lim_{x\to\infty}\frac{3}{\sqrt{x}}.
$$
As $$x\to\infty$$, the denominator $$\sqrt{x}$$ increases without bound, so:
$$
\lim_{x\to\infty}\frac{3}{\sqrt{x}} = 0.
$$

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**24.** Compute  
$$
\lim_{x\to\infty}\frac{4-3x^3}{x^3-1}.
$$
Divide numerator and denominator by $$x^3$$:
$$
\frac{\frac{4}{x^3}-3}{1-\frac{1}{x^3}}.
$$
As $$x\to\infty$$, both $$\frac{4}{x^3}$$ and $$\frac{1}{x^3}$$ approach 0, giving:
$$
\lim_{x\to\infty}\frac{4-3x^3}{x^3-1} = \frac{-3}{1} = -3.
$$

---

**26.** Compute  
$$
\lim_{x\to\infty}\frac{(x-1)^2}{x^3+2}.
$$
Expand the numerator:
$$
(x-1)^2 = x^2 - 2x + 1.
$$
The leading term in the numerator is $$x^2$$ and the leading term in the denominator is $$x^3$$. Dividing numerator and denominator by $$x^3$$:
$$
\frac{x^2 - 2x + 1}{x^3+2} = \frac{\frac{1}{x} - \frac{2}{x^2} + \frac{1}{x^3}}{1+\frac{2}{x^3}}.
$$
As $$x\to\infty$$, all terms with $$x$$ in the denominator approach 0:
$$
\lim_{x\to\infty}\frac{(x-1)^2}{x^3+2} = 0.
$$

---

Thus, the limits are:

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

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