Mock Test 2 — Limits and Continuity

The value of the limit $\lim_{x \to 0} \frac{1 - \cos(2x) \cdot \cos(3x)}{x^2}$ is equal to:

If $f(x) = \lim_{n \to \infty} \frac{x^{2n} - 1}{x^{2n} + 1}$, then the number of points of discontinuity for $f(x)$ across the entire real line is:

Evaluate the limit $\lim_{x \to 0} \frac{\sin x - x \cos x}{x^2 \ln(1+x)}$ using Taylor series expansions. The calculation yields:

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

If the function $f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x} & \text{if } x < 0 \\ c & \text{if } x = 0 \\ \frac{\sqrt{x + bx^2} - \sqrt{x}}{b x^{3/2}} & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$, then the values of the constants are:

The value of the limit $\lim_{x \to \infty} \left( \frac{x^2 + 5x + 3}{x^2 + 1x + 2} \right)^x$ is:

The value of the limit $\lim_{x \to 0} \frac{\cos(\sin x) - 1}{x^2}$ is:

Let $f(x) = [x^2 - 1]$, where $[\cdot]$ denotes the greatest integer function. The number of points of discontinuity for $f(x)$ on the open interval $(1, 2)$ is:

If $\lim_{x \to 0} \frac{a e^x - b \cos x + c e^{-x}}{x \sin x} = 2$, then the values of the constants $a, b, c$ are:

The value of the limit $\lim_{n \to \infty} \frac{[x] + [2x] + [3x] + \dots + [nx]}{n^2}$, where $[\cdot]$ is the floor function, is:

Find the value of the positive constant $a$ if the limit expression satisfies $\lim_{x \to 0} \frac{\ln(1 + ax)}{e^{2x} - 1} = 3$.

Calculate the total number of points of discontinuity across the entire real line for the composite piecewise function $f(x) = \frac{1}{x - [x]}$.

If the jump height of the function $f(x) = \begin{cases} x^2 + c & \text{if } x < 2 \\ 2x + 5 & \text{if } x \ge 2 \end{cases}$ at $x = 2$ is exactly 0, find the value of the constant $c$.

Assertion (A): The function $f(x) = \sin\left(\frac{1}{x}\right)$ features an oscillatory discontinuity at $x = 0$, meaning the limit cannot be resolved to a single value as $x \to 0$.
Reason (R): As $x$ approaches 0, the argument $\frac{1}{x}$ approaches infinity, causing the sine outputs to oscillate infinitely fast between $-1$ and $1$.

Assertion (A): If $f(x)$ and $g(x)$ are two discontinuous functions at $x = a$, then their product function $h(x) = f(x) \cdot g(x)$ must also be discontinuous at $x = a$.
Reason (R): The product of two non-zero limits is indeterminate if one of the individual components does not exist.