📝 Practice Problems

Solution:
Direct substitution: $2(3)^2 - 5(3) + 1 = 2(9) - 15 + 1 = 18 - 15 + 1 = 4$.

Solution:
Factor numerator: $\frac{(x-2)(x+2)}{x-2} = x+2$ for $x \neq 2$.
Then $\displaystyle \lim_{x \to 2} (x+2) = 4$.

Solution:
Multiply numerator and denominator by $\sqrt{x} + 2$:
$\frac{(\sqrt{x} - 2)(\sqrt{x} + 2)}{(x-4)(\sqrt{x} + 2)} = \frac{x - 4}{(x-4)(\sqrt{x} + 2)} = \frac{1}{\sqrt{x} + 2}$ for $x \neq 4$.
Then $\displaystyle \lim_{x \to 4} \frac{1}{\sqrt{x} + 2} = \frac{1}{\sqrt{4} + 2} = \frac{1}{2 + 2} = \frac{1}{4}$.

Solution:
As $x \to 1^-$, $x$ approaches 1 from values less than 1 (like 0.9, 0.99, 0.999).
For these values, $\lfloor x \rfloor = 0$. Therefore, $\displaystyle \lim_{x \to 1^-} \lfloor x \rfloor = 0$.

Solution:
As $x \to 1^+$, $x$ approaches 1 from values greater than 1 (like 1.1, 1.01, 1.001).
For these values, $\lfloor x \rfloor = 1$. Therefore, $\displaystyle \lim_{x \to 1^+} \lfloor x \rfloor = 1$.

Solution:
$\lim_{x \to 1^-} \lfloor x \rfloor = 0$ and $\lim_{x \to 1^+} \lfloor x \rfloor = 1$.
Since the left and right limits are different, the two-sided limit does not exist (DNE).

Solution:
Degrees are equal (both 2). The limit is the ratio of leading coefficients: $\frac{5}{1} = 5$.

Solution:
Degree of numerator (1) < degree of denominator (2). Therefore, the limit is $0$.

Solution:
As $x \to 2$, $(x-2)^2 \to 0^+$ (always positive).
Therefore, $\frac{1}{(x-2)^2} \to \infty$. So the limit is $\infty$.

Solution:
As $x \to 1^+$, denominator $\to 0^+$ → $\frac{3}{0^+} = \infty$.
As $x \to 1^-$, denominator $\to 0^-$ → $\frac{3}{0^-} = -\infty$.
Since one-sided limits differ, the two-sided limit does not exist (DNE).

Solution:
$f(2) = 4$ is defined. $\lim_{x \to 2} x^2 = 4$. Since $\lim = f(2)$, $f$ is continuous at $x = 2$.

Solution:
$f(1)$ is undefined. $\lim_{x \to 1} f(x) = 2$ (exists).
This is a removable discontinuity (hole).

Solution:
$\lim_{x \to 0^-} f(x) = 1$, $\lim_{x \to 0^+} f(x) = -1$.
Since left and right limits exist but are different, this is a jump discontinuity.

Solution:
$f(1) = 1 - 2 - 1 = -2$ (negative). $f(2) = 8 - 4 - 1 = 3$ (positive).
$f$ is continuous on $[1,2]$ and changes sign. By IVT, there exists $c \in (1,2)$ with $f(c) = 0$.

Solution:
$f(1) = 1$, $f(2) = 4$. $3$ is between $1$ and $4$. $f$ is continuous on $[1,2]$.
By IVT, there exists $c \in (1,2)$ with $f(c) = 3$. (Indeed, $c = \sqrt{3} \approx 1.732$.)

Solution:
$-1 \le \cos(1/x) \le 1$ → $-x^2 \le x^2 \cos(1/x) \le x^2$.
$\lim_{x \to 0} -x^2 = 0$ and $\lim_{x \to 0} x^2 = 0$.
By Squeeze Theorem, $\displaystyle \lim_{x \to 0} x^2 \cos(1/x) = 0$.

Solution:
$-1 \le \cos x \le 1$ → $-\frac{1}{x} \le \frac{\cos x}{x} \le \frac{1}{x}$ for $x > 0$.
$\lim_{x \to \infty} -\frac{1}{x} = 0$ and $\lim_{x \to \infty} \frac{1}{x} = 0$.
By Squeeze Theorem, $\displaystyle \lim_{x \to \infty} \frac{\cos x}{x} = 0$.

Solution:
Divide numerator and denominator by $x^3$: $\frac{2 - \frac{1}{x^2}}{1 + \frac{4}{x^3}}$.
As $x \to -\infty$, $\frac{1}{x^2} \to 0$ and $\frac{4}{x^3} \to 0$.
So $\displaystyle \lim_{x \to -\infty} f(x) = \frac{2}{1} = 2$.

Solution:
$f$ is a rational function, continuous on its domain $(-\infty,3) \cup (3,\infty)$.
Since $x = 3$ is in $[0,5]$ and $f$ is undefined there, $f$ is not continuous on $[0,5]$.

Solution:
By IVT, $f$ takes every value between $f(0)=1$ and $f(4)=5$.
$2$ is between $1$ and $5$, so $f$ must take the value $2$.
$0$ is below $1$, $6$ is above $5$ — IVT does not guarantee those values.
Answer: (B) 2.