Mixed review R1: notation to tangent lines

The input change has a sign

The new input minus the old input is $(x-h)-x=-h$. Substitute the whole new input:

Since $h\ne0$, cancel the nonzero factor to get $\boxed{4x-2h}$. Using $h$ instead would reverse the sign of the rate relative to the stated input change. The limiting derivative is $4x$, but the expression asked for retains the interval term $-2h$.

A denominator square root must be strictly positive

The logarithm requires $9-x^2>0$, or $-3<x<3$. The denominator requires $x+1>0$: equality would make the denominator zero. Intersecting gives $\boxed{(-1,3)}$. It is not enough to check only that the square root is real.

A matching limit is still not a matching point value

The left limit is $k$ and the right limit is $1$. Therefore the two-sided limit exists precisely when $k=1$, and its value is then $1$. But $g(0)=2$, so no $k$ makes the function continuous at zero. For $k\ne1$ there is no two-sided limit; for $k=1$ there is a removable mismatch with the point value.

Expose a factor by a conjugate identity

For nearby $x\ne0$, multiply by the conjugate over itself:

The limit is $\boxed{1/4}$ by continuity of the last expression near zero. Initially there was no visible common factor $x$ multiplying the entire numerator; a term inside a square root cannot be deleted as though it were such a factor.

Three different objects from the same position function

The endpoint positions are $s(1)=0$ and $s(3)=6$, so average velocity is $(6-0)/(3-1)=\boxed3$ m/s. At $t=1$,

Its limit is $s'(1)=\boxed1$ m/s. The tangent line goes through $(1,0)$ with slope $1$, giving $\boxed{s=t-1}$ as the tangent-line equation. The secant’s slope $3$ and tangent’s slope $1$ are not competing answers; they describe different rates.

Finite sampling suggests a limit but does not establish it

The measurements suggest $\boxed{f'(2)=3}$, because both signs of small $h$ give quotients near $3$. They do not constrain every arbitrarily small nonzero input. For instance, a quotient rule could equal $3+h$ at these four sampled values but equal $3+\sin(1/h)$ at other nonzero inputs. Setting $f(2)=0$ and $f(2+h)=hQ(h)$ produces a function consistent with the samples but with no derivative at $2$. A formula, bound, or argument controlling all sufficiently small $h$ is needed for a proof.

A continuous corner has two unequal limiting slopes

Let $a=-1/2$. Then $f(a)=0$ and $f(a+h)=|2h|=2|h|$. The quotient is $2|h|/h$, equal to $2$ for $h>0$ and $-2$ for $h<0$. Thus the two-sided derivative does not exist. Continuity does hold because $|f(a+h)-f(a)|=2|h|\to0$. A graph can have no break in height and still have a corner.

A finite limiting rate forces the height change to vanish

For $h\ne0$,

The first factor tends to the finite number $f'(a)$ and the second tends to zero. The product limit is therefore $f'(a)\cdot0=0$, so $f(a+h)\to f(a)$, which is continuity. Finiteness permits the ordinary product-limit law; treating an infinite quotient as an ordinary real number and multiplying it by zero would not justify this step. The converse fails, as R1.07 demonstrates.