1.8. $$ \lim _{x \rightarrow 1}\left(\frac{x^{4}-1}{x^{3}-1}\right) $$ 1.9. $$ \lim _{x \rightarrow 4}\left(\frac{\sqrt{5-x}-1}{2-\sqrt{x}}\right) $$ 1.10. $$ \lim _{x \rightarrow 1}\left(\frac{x^{4}-1}{1-\sqrt{x}}\right) $$ 1.11. $$ \lim _{x \rightarrow 25}\left(\frac{\frac{1}{\sqrt{x}}-\frac{1}{5}}{x-25}\right) $$

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**1.8.**  
We want to evaluate  
$$
\lim_{x \rightarrow 1} \frac{x^4-1}{x^3-1}.
$$  
Factor the numerator and denominator:
$$
x^4-1 = (x-1)(x^3+x^2+x+1)
$$
$$
x^3-1 = (x-1)(x^2+x+1).
$$
For $$x\neq 1$$, we cancel the common factor $$x-1$$ to obtain
$$
\frac{x^4-1}{x^3-1} = \frac{x^3+x^2+x+1}{x^2+x+1}.
$$
Now substitute $$x=1$$:
$$
\frac{1^3+1^2+1+1}{1^2+1+1} = \frac{4}{3}.
$$

---

**1.9.**  
We wish to compute  
$$
\lim_{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}}.
$$
Substitute $$x=4$$:
$$
\sqrt{5-4}-1 = \sqrt{1}-1 = 0,\quad \text{and} \quad 2-\sqrt{4} = 2-2 = 0.
$$
Since this is a $$\frac{0}{0}$$ indeterminate form, we apply L’Hôpital’s rule by differentiating the numerator and the denominator.

Differentiate the numerator:
$$
\frac{d}{dx} \left(\sqrt{5-x}-1\right) = \frac{d}{dx}\left((5-x)^{1/2}\right) = -\frac{1}{2}(5-x)^{-1/2}.
$$
Differentiate the denominator:
$$
\frac{d}{dx} \left(2-\sqrt{x}\right) = -\frac{1}{2}x^{-1/2}.
$$
Thus, the limit becomes
$$
\lim_{x \rightarrow 4} \frac{-\frac{1}{2}(5-x)^{-1/2}}{-\frac{1}{2}x^{-1/2}} = \lim_{x \rightarrow 4} \frac{x^{1/2}}{(5-x)^{1/2}}.
$$
Now substitute $$x=4$$:
$$
\frac{\sqrt{4}}{\sqrt{5-4}} = \frac{2}{1} = 2.
$$

---

**1.10.**  
We wish to evaluate  
$$
\lim_{x \rightarrow 1} \frac{x^4-1}{1-\sqrt{x}}.
$$
Direct substitution gives $$0/0$$, so we simplify. First, factor the numerator:
$$
x^4-1 = (x-1)(x^3+x^2+x+1).
$$
Next, express $$x-1$$ in terms of $$\sqrt{x}$$:
$$
x-1 = (\sqrt{x})^2-1 = (\sqrt{x}-1)(\sqrt{x}+1).
$$
Thus, we can rewrite the numerator as
$$
x^4-1 = (\sqrt{x}-1)(\sqrt{x}+1)(x^3+x^2+x+1).
$$
Notice that the denominator can be written as
$$
1-\sqrt{x} = -(\sqrt{x}-1).
$$
Now, cancel the common factor $$\sqrt{x}-1$$ (for $$x\neq 1$$):
$$
\frac{x^4-1}{1-\sqrt{x}} = \frac{(\sqrt{x}-1)(\sqrt{x}+1)(x^3+x^2+x+1)}{- (\sqrt{x}-1)} = -(\sqrt{x}+1)(x^3+x^2+x+1).
$$
Substitute $$x=1$$:
$$
-(\sqrt{1}+1)(1^3+1^2+1+1) = -(1+1)(4) = -8.
$$

---

**1.11.**  
We wish to determine  
$$
\lim_{x \rightarrow 25} \frac{\frac{1}{\sqrt{x}}-\frac{1}{5}}{x-25}.
$$
First, combine the fractions in the numerator:
$$
\frac{1}{\sqrt{x}}-\frac{1}{5} = \frac{5-\sqrt{x}}{5\sqrt{x}}.
$$
Thus, the limit becomes
$$
\lim_{x \rightarrow 25} \frac{5-\sqrt{x}}{5\sqrt{x}(x-25)}.
$$
Notice that the term $$5-\sqrt{x}$$ can be related to $$x-25$$. Observe that
$$
x-25 = (\sqrt{x})^2-5^2 = (\sqrt{x}-5)(\sqrt{x}+5) = - (5-\sqrt{x})(\sqrt{x}+5).
$$
Therefore,
$$
\frac{5-\sqrt{x}}{x-25} = \frac{5-\sqrt{x}}{- (5-\sqrt{x})(\sqrt{x}+5)} = -\frac{1}{\sqrt{x}+5}.
$$
The limit now becomes
$$
\lim_{x \rightarrow 25} -\frac{1}{5\sqrt{x}(\sqrt{x}+5)}.
$$
Substitute $$x=25$$ (noting $$\sqrt{25}=5$$):
$$
-\frac{1}{5(5)(5+5)} = -\frac{1}{25 \times 10} = -\frac{1}{250}.
$$

---

$$
\boxed{1.8. \, \frac{4}{3}, \quad 1.9. \, 2, \quad 1.10. \, -8, \quad 1.11. \, -\frac{1}{250}}
$$
$$ \lim_{x \rightarrow 1} \frac{x^4-1}{x^3-1} = \frac{4}{3}, \quad \lim_{x \rightarrow 4} \frac{\sqrt{5-x}-1}{2-\sqrt{x}} = 2, \quad \lim_{x \rightarrow 1} \frac{x^4-1}{1-\sqrt{x}} = -8, \quad \lim_{x \rightarrow 25} \frac{\frac{1}{\sqrt{x}}-\frac{1}{5}}{x-25} = -\frac{1}{250} $$

1.8.
1.9.
1.10.
1.11.

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