Continuity and Differentiability: questions with answers

The left hand derivative at $1$ is $-1$ and the right hand derivative is $1$. Both limits exist -- but the derivative is a single two-sided limit, and a two-sided limit exists only when the two one-sided limits agree. They do not, so $f'(1)$ does not exist. The graph has a corner: continuous, but with no single tangent direction.

Theorem 2 asks for $g$ continuous at $c$ and $h$ continuous at $g(c)$ -- the input $h$ actually receives. Here $g(-5) = 4$, and $h$ is continuous there. Testing $h$ at $x = -5$ itself would be testing it at a value it never sees; in fact $h$ is not even continuous at $-5$, and the composite is perfectly continuous all the same.

Yes, exactly one. The quotient reduces to $x + 2$ for $x \neq 2$, so the limit at $2$ exists and equals $4$, and $k = 4$ is the one value that matches it. The $\frac{0}{0}$ from direct substitution does not mean no $k$ works -- it means the expression has to be simplified first. And $k$ is not free: any other value leaves the point out of step with the limit.

Differentiable but not continuous cannot happen: Theorem 3 makes continuity a consequence of differentiability. The other three all occur -- $x^2$ at $0$ (both), $|x|$ at $0$ (continuous, not differentiable), $[x]$ at $1$ (neither).

For any $n \geq 1$ the factor $x^n$ tends to $0$ and the bounded sine factor cannot stop the product tending to $0$, which is $f(0)$. For $n = 0$ there is no damping at all and the function is $\sin \frac{1}{x}$, which has no limit. So the boundary is $n = 1$: one power is already enough, demanding a square asks for damping that is not needed, and no power at all is exactly the case that fails.

Because $0$ is not in the domain. Continuity is asserted or denied only at points where the function is defined, and $f$ is continuous at every one of those. The pen lifts at $0$ because the graph has a hole there -- a gap in the domain, not a point of discontinuity. It is true that the two pieces both approach $2$ at $0$, but that is beside the point: even if they did not, $0$ would still not be a point of discontinuity, because it is not a point of $f$.

The formula comes from dividing the chain rule by $\dfrac{dx}{dt}$, which is allowed only where $\dfrac{dx}{dt} = 2t$ is non-zero. It vanishes at $t = 0$, and there the formula says nothing -- the curve may have a vertical tangent, a cusp, or a perfectly good derivative that must be found another way. Where $\dfrac{dy}{dt}$ vanishes the formula is fine: it simply gives $0$.

The outer derivative is read at the value the outer function actually receives, which is $g(x) = x^2$: $f'(x^2) = \cos(x^2)$. Reading it at $x$ gives $\cos x$, the derivative of a different function altogether. The inner derivative $g'(x) = 2x$ is the other factor, not the argument.

From $\cos y = x$: $-\sin y \, \dfrac{dy}{dx} = 1$, so $\dfrac{dy}{dx} = -\dfrac{1}{\sin y} = -\dfrac{1}{\sqrt{1 - x^2}}$. The minus sign is not decorative: $\sin^{-1} x + \cos^{-1} x = \dfrac{\pi}{2}$ is constant, so the two derivatives must be negatives of each other, not equal. $\cos^{-1} x$ is the inverse function, not $\dfrac{1}{\cos x}$.

$xy^2$ is a product of a function of $x$ and a function of $y$, and both factors vary with $x$. The product rule gives $y^2 + 2xy \dfrac{dy}{dx}$. Treating $x$ as a mere coefficient loses the term from differentiating it; dropping $\dfrac{dy}{dx}$ differentiates the $y$ factor with respect to $y$ rather than $x$.

Each branch is a polynomial, so $f$ is continuous everywhere except possibly at the joint -- and continuity AT the joint is a comparison between the branches, not a property of either one. The left hand limit is $5$ and the right hand limit is $1$; they differ, so $f$ is not continuous at $3$. The value $f(3) = 5$ comes from the branch that carries the equals sign, not from the strict one.

Change of base puts the argument on top: $\dfrac{\log 8}{\log 2} = 3$ is exactly the known fact rewritten. The inverted quotient is the reciprocal, $\dfrac{1}{3}$ or $\dfrac{1}{2}$, not the known value; and neither the logarithm of the quotient nor the difference of the logarithms equals a logarithm to the old base -- checking on a case with a known answer is precisely how to see which arrangement is right.

No. The working took $\log(-2)$, which does not exist, so nothing that follows applies there. The final formula may contain no logarithm of a negative number and look perfectly usable -- the algebra runs through without complaint -- but a derivation that begins with an undefined quantity proves nothing about that point.

First $\dfrac{dy}{dx} = \dfrac{3t}{2}$. Then differentiate that with respect to $t$ and divide by $\dfrac{dx}{dt} = 2t$: $\dfrac{d^2y}{dx^2} = \dfrac{3}{4t}$. Stopping after the $t$-derivative gives $\dfrac{3}{2}$, a rate against the parameter rather than against $x$; $\left(\dfrac{dy}{dx}\right)^2 = \dfrac{9t^2}{4}$ is not a second derivative.

Change of base: $\log_{3} x = \dfrac{\log x}{\log 3}$, and $\dfrac{1}{\log 3}$ is a constant factor. So the derivative is $\dfrac{1}{\log 3} \cdot \dfrac{1}{x} = \dfrac{1}{x \log 3}$. Only the natural logarithm differentiates to $\dfrac{1}{x}$; another base scales it by a constant.