Precalculus Introduction to Limits Worksheet Solutions (PDF)

Precalculus math worksheet solutions: Introduction to Limits – Maths PDF

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This limits practice solutions for Precalculus covers introduction to limits (Common Core F-IF.B.6, F-IF.C.7d), with a free PDF to print.

Use this key to check each limit. Solutions show the table, the algebraic simplification or the degree comparison, with final answers in bold.

Precalculus · Common Core State Standards: F-IF.B.6, F-IF.C.7d

Limits practice solutions: exercises 1 to 10

Exercise 1 – Average rate of change

$\frac{f(3)-f(1)}{3-1}=\frac{9-1}{2}$.

Exercise 2 – Average rate of a cubic

$\frac{8-1}{1}$.

Exercise 3 – Average velocity

$\frac{s(3)-s(1)}{2}=\frac{144-16}{2}$.

$64$ ft/s

Exercise 4 – Shrinking interval

The secant slope approaches the tangent slope 4.

$\frac{(2+h)^{2}-4}{h}=\frac{4h+h^{2}}{h}=4+h$. As $h\to0$ this approaches $4$.

$4+h$; the slope approaches $4$

Exercise 5 – Limit from a table

The values approach 6.

For $x\neq3$, $f(x)=x+3$. The values approach $6$ from both sides.

The limit is 6

Exercise 6 – Direct substitution

The function is a polynomial, so substitute: $12-2+1$.

Exercise 7 – Factor and cancel

$\frac{(x-1)(x+1)}{x+1}=x-1$ for $x\neq-1$. The limit is $-1-1$.

$-2$

Exercise 8 – Rationalize

Multiply by $\frac{\sqrt x+2}{\sqrt x+2}$: $\frac{x-4}{(x-4)(\sqrt x+2)}=\frac{1}{\sqrt x+2}$. At $x=4$: $\frac14$.

$\frac14$

Exercise 9 – Expand first

$(x+5)^{2}-25=x^{2}+10x$, so the quotient is $x+10$ for $x\neq0$.

Exercise 10 – One-sided limits from a graph

Left limit 3; right limit 4; f(2) = 4.

The left piece heads to height $3$ (open circle). The right piece starts at height $4$ (closed point). The one-sided limits differ.

$3,\ 4,\ f(2)=4$; the limit does not exist

Exercises 11 and up

Exercise 11 – A piecewise limit

Left: $2(3)-1=5$. Right: $3+2=5$. $f(3)=5$.

Both limits are 5 and f(3) = 5, so it is continuous

Exercise 12 – Limit at infinity (equal degrees)

The graph levels off at y = 2.

Divide numerator and denominator by $x$: $\frac{2+\frac1x}{1-\frac1x}\to\frac{2}{1}$.

Exercise 13 – Limit at infinity (higher powers)

Divide by $x^{2}$: $\frac{3-\frac5x}{1+\frac{4}{x^{2}}}\to3$.

Exercise 14 – Limit at infinity (lower numerator)

The denominator grows faster. Dividing by $x^{2}$: $\frac{\frac4x+\frac{1}{x^{2}}}{1+\frac{2}{x^{2}}}\to0$.

Exercise 15 – Infinite one-sided limits

For $x$ slightly above 2, $x-2$ is a tiny positive number, so the quotient is huge and positive. Slightly below 2 it is huge and negative.

$\infty$ and $-\infty$; the two-sided limit does not exist

Exercise 16 – Removable discontinuity

The point (2, 5) sits above the hole at (2, 4).

$\lim_{x\to2}f(x)=4$ but $f(2)=5$, so it is not continuous. Redefine $f(2)=4$ to fill the hole.

Not continuous; set f(2) = 4

Exercise 17 – Make a function continuous

Left limit: $4+1=5$. Right: $2k-1$. Set equal: $2k-1=5$.

$k=3$

Exercise 18 – Delivery fee

Closed circle at the right end of each step.

Just below 1 lb the fee is $\$5$; just above it is $\$8$. The limit does not exist at $w=1$.

Left $\$5$; right $\$8$; no two-sided limit

Exercise 19 – Instantaneous velocity

$\frac{(3+h)^{2}+2(3+h)-15}{h}=\frac{8h+h^{2}}{h}=8+h$. As $h\to0$ the limit is $8$. Check: $h=0.1$ gives $8.1$; $h=0.01$ gives $8.01$.

$8+h$; $8$ ft/s

Exercise 20 – Drug concentration

The curve approaches 5 mg/L.

$C(8)=\frac{40}{10}=4$. Equal degrees: the limit is $\frac51=5$. The concentration approaches $5$ mg/L but never exceeds it.

$4$ mg/L; $5$ mg/L

Key facts behind this limits practice solutions

Key facts

Average rate of change: (f(b) – f(a))/(b – a), the secant slope.
lim x to a of f(x) = L means f(x) approaches L as x approaches a from both sides.
The limit can exist even if f(a) is undefined (a hole).
Two-sided limit exists only if the left and right limits are equal.
f is continuous at a if f(a) exists, the limit exists and they are equal.

How to use this limits practice solutions

Cover each solution, solve the exercise again, then uncover it and compare every line. If your answer differs, find the first step that changed and redo the exercise from there.

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