Mock Test 1 — Limits and Continuity

The value of the limit $\lim_{x \to 0} \frac{\tan(3x)}{\ln(1+2x)}$ is equal to:

If $f(x) = \begin{cases} \frac{x^2 - 1}{x - 1} & \text{if } x \neq 1 \\ c & \text{if } x = 1 \end{cases}$ is continuous at $x = 1$, then the constant $c$ is:

Evaluate the value of the algebraic limit $\lim_{x \to 2} \frac{x^5 - 32}{x - 2}$. The calculation yields:

The value of the limit $\lim_{x \to 0} \frac{1 - \cos x}{x \sin x}$ is:

The function $f(x) = \frac{x}{|x|}$ features which specific type of discontinuity at $x = 0$?

The value of the limit $\lim_{x \to \infty} \frac{3x^3 - 4x^2 + 5}{2x^3 + 7x - 1}$ is:

Evaluate $\lim_{x \to 0} (1 + 2x)^{\frac{1}{x}}$. The series converges to:

The number of points of discontinuity for the function $f(x) = \frac{1}{\sin x}$ on the open interval $(0, 2\pi)$ is:

The value of the limit $\lim_{x \to 0} x \sin\left(\frac{1}{x}\right)$ evaluated using the Sandwich Theorem is:

If a continuous function satisfies $f(1) = -5$ and $f(4) = 5$, then the Intermediate Value Theorem guarantees that the curve crosses the x-axis inside the interval $(1, 4)$ at least:

Find the value of the constant $a$ that makes the function $f(x) = \begin{cases} a x + 3 & \text{if } x \le 2 \\ x^2 - 1 & \text{if } x > 2 \end{cases}$ continuous at the breakpoint $x = 2$.

Calculate the jump height of the piecewise function $f(x) = \begin{cases} 2x + 1 & \text{if } x < 1 \\ 5x - 1 & \text{if } x \ge 1 \end{cases}$ at $x = 1$.

Find the value of the limit exponent parameter $k$ if $\lim_{x \to 0} \frac{e^{5x} - 1}{x} = k$.

Assertion (A): For a function $f(x)$, if the left-hand limit equals 2 and the right-hand limit equals 2, the two-sided limit is guaranteed to exist and equal 2.
Reason (R): A two-sided limit exists at a point if and only if both one-sided limits exist, are finite, and are identically equal.

Assertion (A): The function $f(x) = \frac{1}{x}$ is continuous on the open interval $(0, 1)$, but features an infinite discontinuity at the boundary point $x = 0$.
Reason (R): As $x \to 0^+$, the function values approach positive infinity ($\lim_{x \to 0^+} \frac{1}{x} = \infty$), creating a vertical asymptote at the boundary.