1.19. $$ \lim _{x \rightarrow 4}\left[\frac{4 x-x^{2}}{2-\sqrt{x}}\right] $$ 1.20. $$ \lim _{t \rightarrow 2} \frac{2^{2 t}+2^{t}-20}{2^{t}-4} $$

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**Problem 1.19**

We need to evaluate:
$$
\lim_{x \rightarrow 4} \frac{4x - x^2}{2-\sqrt{x}}
$$

1. Write the numerator in factored form:
   $$
   4x-x^2 = x(4-x)
   $$
2. Introduce the substitution:
   $$
   u = \sqrt{x} \quad \Longrightarrow \quad x = u^2.
   $$
   Then, the limit as $$ x \to 4 $$ corresponds to $$ u \to 2 $$.

3. Rewrite the numerator in terms of $$ u $$:
   $$
   4x - x^2 = 4u^2 - u^4 = u^2(4-u^2).
   $$

4. Notice that the expression $$ 4-u^2 $$ factors further:
   $$
   4-u^2 = (2-u)(2+u).
   $$

5. Now, substitute these into the limit:
   $$
   \frac{4x-x^2}{2-\sqrt{x}} = \frac{u^2 (2-u)(2+u)}{2-u}.
   $$

6. Cancel the common factor $$(2-u)$$ (valid for $$ u \neq 2 $$):
   $$
   \frac{u^2 (2-u)(2+u)}{2-u} = u^2 (2+u).
   $$

7. Take the limit as $$ u \to 2 $$:
   $$
   \lim_{u \rightarrow 2} u^2 (2+u) = 2^2 (2+2) = 4 \times 4 = 16.
   $$

Thus,
$$
\lim_{x \rightarrow 4} \frac{4x-x^2}{2-\sqrt{x}} = 16.
$$

---

**Problem 1.20**

We need to evaluate:
$$
\lim_{t \rightarrow 2} \frac{2^{2t}+2^t-20}{2^t-4}
$$

1. Define the substitution:
   $$
   u = 2^t.
   $$
   As $$ t \to 2 $$, then $$ u \to 2^2 = 4 $$.

2. Rewrite the expression in terms of $$ u $$. Note that:
   $$
   2^{2t} = (2^t)^2 = u^2.
   $$
   So the expression becomes:
   $$
   \frac{u^2 + u - 20}{u-4}.
   $$

3. Factor the numerator:
   $$
   u^2 + u - 20 = (u + 5)(u - 4).
   $$

4. Cancel the common factor $$(u-4)$$ (valid for $$ u \neq 4 $$):
   $$
   \frac{(u+5)(u-4)}{u-4} = u+5.
   $$

5. Now, take the limit as $$ u \to 4 $$:
   $$
   \lim_{u \to 4} (u+5) = 4+5 = 9.
   $$

Thus,
$$
\lim_{t \rightarrow 2} \frac{2^{2t}+2^t-20}{2^t-4} = 9.
$$
**Problem 1.19** $$ \lim_{x \rightarrow 4} \frac{4x - x^2}{2 - \sqrt{x}} = 16 $$ **Problem 1.20** $$ \lim_{t \rightarrow 2} \frac{2^{2t} + 2^t - 20}{2^t - 4} = 9 $$

1.19.
1.20.

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