Differential Calculus Questions and Answers

Answer: A

Write the quotient so that the standard limit appears with a matching argument:

\[ \frac{\sin 5x}{3x} \;=\; \frac{5}{3}\cdot\frac{\sin 5x}{5x}. \]

Since \( \sin\theta/\theta \to 1 \) as \( \theta \to 0 \), the second factor tends to one and

\[ \lim_{x \to 0} \frac{\sin 5x}{3x} \;=\; \frac{5}{3}. \]

The reciprocal \( 3/5 \) comes of matching the arguments the wrong way round

Answer: D

Use the half-angle identity \( 1 - \cos x = 2\sin^{2}(x/2) \):

\[ \frac{1-\cos x}{x^{2}} \;=\; \frac{2\sin^{2}(x/2)}{x^{2}} \;=\; \frac{1}{2}\left(\frac{\sin (x/2)}{x/2}\right)^{2}. \]

The bracket tends to one, so the limit is \( 1/2 \). Two applications of L'Hopital's rule give the same value, the numerator differentiating to \( \sin x \) and then to \( \cos x \)

Answer: C

Take logarithms and let \( L \) be the limit:

\[ \ln L \;=\; \lim_{x \to \infty} 2x \ln\!\left(1+\frac{3}{x}\right) \;=\; \lim_{x \to \infty} 2x \cdot \frac{3}{x} \;=\; 6, \]

using \( \ln(1+t) \approx t \) for small \( t \). Hence \( L = e^{6} \). The general rule is that \( (1+a/x)^{bx} \to e^{ab} \), and here \( ab = 3 \times 2 \)

Answer: B

Replace each factor by its leading term, \( e^{2x} - 1 \approx 2x \) and \( \tan 3x \approx 3x \):

\[ \lim_{x \to 0}\frac{e^{2x}-1}{\tan 3x} \;=\; \lim_{x \to 0}\frac{2x}{3x} \;=\; \frac{2}{3}. \]

L'Hopital's rule gives the same result, the derivatives being \( 2e^{2x} \) and \( 3\sec^{2}3x \), both equal to their coefficients at the origin

Answer: A

For \( x > 0 \) the quotient is \( x/x = 1 \), and for \( x < 0 \) it is \( -x/x = -1 \). So

\[ \lim_{x \to 0^{+}} f(x) = 1, \qquad \lim_{x \to 0^{-}} f(x) = -1. \]

The two disagree, so the two-sided limit does not exist and the origin carries a jump discontinuity of magnitude two

Answer: D

A removable discontinuity is one that redefining a single value repairs: the limit \( \lim_{x \to a} f(x) \) exists, but \( f(a) \) is either absent or set to something else. Unequal one-sided limits give a jump, unbounded growth an infinite discontinuity, and faster oscillation an essential one, and none of those three can be repaired by changing one value

Answer: C

Cancel the common factor away from the point itself:

\[ \frac{x^{2}-9}{x-3} \;=\; \frac{(x-3)(x+3)}{x-3} \;=\; x+3 \quad (x \neq 3). \]

So \( \lim_{x \to 3} f(x) = 6 \), and assigning \( f(3) = 6 \) removes the discontinuity. The quotient itself is \( 0/0 \) at the point, which is why the value must be supplied

Answer: B

Differentiability implies continuity but not the reverse. The function \( |x| \) is continuous everywhere yet has no derivative at the origin, where the one-sided slopes are \( -1 \) and \( +1 \). So continuity is the weaker condition, and a corner is exactly the case that separates the two properties

Answer: A

Rolle's theorem asks for \( f(a) = f(b) \), and concludes that \( f'(c) = 0 \) for some \( c \) between them. Nothing is asked of the sign of the derivative, of monotonicity, or of where the interval sits: dropping the endpoint condition gives Lagrange's theorem instead, whose conclusion is a slope rather than a zero

Answer: D

The endpoint values agree, \( f(1) = f(3) = 0 \), so the theorem applies. Setting the derivative to zero,

\[ f'(x) \;=\; 2x - 4 \;=\; 0 \quad \Longrightarrow \quad c \;=\; 2, \]

which lies inside the interval as the theorem promises. It is the vertex of the parabola, the one place where the tangent is horizontal