Limits Practice

These questions follow The Limit and The Derivative — watching limits happen numerically, taking them algebraically, and reading what a difference quotient says about a graph. Take the numeric road first where you can; the estimate audits the algebra.

Watching limits happen

1. For $f(x)=x^2$, compute the average rate of change from $x=3$ to $x=3+h$ for $h=1$, $h=0.1$, and $h=0.01$. What limit are the answers approaching?
2. Evaluate ${\left(1+\frac{1}{n}\right)}^n$ for $n=1$, $2$, $3$, and $10$. What number is this sequence creeping toward?
3. The Fibonacci sequence begins $1,1,2,3,5,8,13,\dots$ Compute the ratio of each term to the one before it, out to $\frac{13}{8}$. What value do the ratios approach?

Answer 1

Each average rate is $\frac{(3+h{)}^2-9}{h}$. For $h=1$: $\frac{16-9}{1}=7$. For $h=0.1$: $\frac{9.61-9}{0.1}=6.1$. For $h=0.01$: $\frac{9.0601-9}{0.01}=6.01$. The march is $7$, $6.1$, $6.01$ — settling toward $6$, the slope of the tangent to $y=x^2$ at $x=3$.

Answer 2

$n=1$: $2$. $n=2$: $1.5^2=2.25$. $n=3$: ${\left(\frac{4}{3}\right)}^3=\frac{64}{27}\approx 2.370$. $n=10$: $1.1^{10}\approx 2.594$. The sequence climbs toward $e\approx 2.718$ — slowly, but it never stops climbing and never passes $e$. You will meet this number properly in Derivatives of Exponential Functions.

Answer 3

$\frac{1}{1}=1$, $\frac{2}{1}=2$, $\frac{3}{2}=1.5$, $\frac{5}{3}\approx 1.667$, $\frac{8}{5}=1.6$, $\frac{13}{8}=1.625$. The ratios bounce alternately above and below their destination, closing in on the golden ratio $\frac{1+\sqrt{5}}{2}\approx 1.618$. A limit reached from both sides at once.

Taking limits algebraically

4. Evaluate $\underset{x\to 2}{\lim }\frac{x^2-4}{x-2}$.
5. Evaluate $\underset{h\to 0}{\lim }\frac{(3+h{)}^2-9}{h}$ by simplifying first, and confirm it matches question 1.
6. Evaluate $\underset{h\to 0}{\lim }\frac{(x+h{)}^3-x^3}{h}$.
7. Evaluate $\underset{x\to 0}{\lim }\frac{\sqrt{x+9}-3}{x}$. (Multiplying by a well-chosen form of 1 helps.)

Answer 4

Substituting $x=2$ gives $\frac{0}{0}$ — no verdict. Factor: $\frac{(x-2)(x+2)}{x-2}=x+2$ everywhere except $x=2$ itself, and the limit only cares about the approach. As $x\to 2$, $x+2\to 4$. The limit is $4$.

Answer 5

Expand: $\frac{9+6h+h^2-9}{h}=\frac{6h+h^2}{h}=6+h$. As $h\to 0$ this heads straight for $6$ — the same destination the numeric march in question 1 was pointing at. Two roads, one limit.

Answer 6

Expand the cube: $(x+h{)}^3=x^3+3x^{2}h+3x{h}^2+h^3$, so the quotient is $\frac{3x^{2}h+3x{h}^2+h^3}{h}=3x^2+3xh+h^2$. As $h\to 0$, the limit is $3x^2$ — which is exactly the derivative of $x^3$, computed from the definition.

Answer 7

Multiply numerator and denominator by $\sqrt{x+9}+3$: $\frac{(x+9)-9}{x(\sqrt{x+9}+3)}=\frac{1}{\sqrt{x+9}+3}$ As $x\to 0$ this becomes $\frac{1}{3+3}=\frac{1}{6}$. Numeric audit: at $x=0.01$, $\frac{\sqrt{9.01}-3}{0.01}\approx 0.1666$. ✓

Reading the derivative as a limit

8. You are told $\underset{h\to 0}{\lim }\frac{f(4+h)-f(4)}{h}=8$ for $f(x)=x^2$. What does this say about the graph of $f$? What would the same statement mean for a general function $y=f(x)$?
9. Use the definition of the derivative to determine $f^{\prime }(x)$ for $f(x)=x^2-5x$.

Answer 8

That limit is the definition of $f^{\prime }(4)$: the slope of the tangent to $y=x^2$ at the point $(4,16)$ is $8$ — equivalently, the instantaneous rate of change there is 8 units of $y$ per unit of $x$. For a general $f$, the statement says exactly the same thing about the point $(4,f(4))$: tangent slope 8, whatever the function is. The sentence is about the point, not the formula.

Answer 9

$\begin{array}{rl}\frac{(x+h{)}^2-5(x+h)-(x^2-5x)}{h}& \\ & =\frac{2xh+h^2-5h}{h}=2x+h-5\end{array}$ As $h\to 0$: $f^{\prime }(x)=2x-5$. Sanity check with the toolbox to come: the power rule will say the same thing in one line — but you just proved it, and that is better.