Limits, Continuity and Differentiability Questions and Answers

Answer: D

c is 5.

The line joining (2, 0) and (8, 0) is horizontal, with slope 0. So we need \(f'(c) = 0\).

\[ f'(c) = 2c - 10 = 0 \Rightarrow c = 5 \]

This is the mean value theorem: 2 and 8 are the roots of f, and 5, the midpoint, is where the parabola's tangent is horizontal. A sanity check: the vertex of the parabola \(x^{2} - 10x + 16\) is at \(x = -\frac{-10}{2} = 5\), the only place its tangent is horizontal.

Answer: E

The limit is sin1.

As \(x \to 0^{-}\), \(x\) is a small negative number, so \([x] = -1\) throughout the approach. The expression is constant there:

\[ \begin{aligned} \frac{\sin[x]}{[x]} &= \frac{\sin(-1)}{-1} \\ &= \frac{-\sin 1}{-1} \\ &= \sin 1 \end{aligned} \]

The trap is to use \(\frac{\sin\theta}{\theta} \to 1\), which needs \(\theta \to 0\); here \([x]\) stays at -1 and never approaches 0. Forgetting that the two minus signs cancel gives -sin1.

Answer: A

|x| is continuous but not differentiable at x = 0.

It is continuous because |x| approaches 0 from both sides and f(0) = 0. But its graph has a sharp corner at the origin: the slope is -1 to the left and +1 to the right, so

\[ \begin{aligned} \lim_{h \to 0^{-}} \frac{|h|}{h} &= -1 \\ \lim_{h \to 0^{+}} \frac{|h|}{h} &= 1 \end{aligned} \]

The one-sided derivatives differ, so there is no derivative at 0. A function differentiable at a point must be continuous there, which rules out option B.

Answer: B

The derivative is \(x^{x}(1 + \ln x)\).

Neither the power rule nor the exponential rule applies, because both the base and the exponent vary. Take logarithms first:

\[ \begin{aligned} y &= x^{x} \\ \ln y &= x\ln x \\ \frac{1}{y}\frac{dy}{dx} &= \ln x + 1 \\ \frac{dy}{dx} &= x^{x}(1 + \ln x) \end{aligned} \]

x^x-1 wrongly treats the exponent as a constant (power rule), x ln x is ln y itself, and ln x is only part of the derivative of x ln x.