cal1-asgn1

Calculus I Assignment # 1

This assignment covers

A: Algebraic and Trigonometric Limits

B: Limits on Restricted Domains

C: Continuity

QUESTIONS

A) Evaluate These Limits

If the limit does not exist, state why.

1) 2) 3)

4) 5) 6)

B) Limits on Restricted Domains (piecewise defined functions)

Discuss restrictions on the domains and the limits for these functions.

1) 2)

3)

C) Continuity

1) Discuss the discontinuities of these functions:
For removeable jumps, find the limit.

1) 2)

3) 4)

2) Find values for c and/or c and d so that f(x) is continuous for all Real numbers.

1) 2)

3) 4)

SOLUTIONS

A) Evaluate These Limits

If the limit does not exist, state why.

1) 2)

3) 4)

5) 6)

B) Limits on Restricted Domains (piecewise defined functions)

Discuss restrictions on the domains and the limits for these functions.

1)

so limit = 7
2)

so limit = 1

3)
therefore, limit doesn't exist.

C) Continuity

1) Discuss the discontinuities of these functions:
For removeable jumps, find the limit.

1) is restricted to -3 [ x [ 3
and x⁴ - 16 has vertical asymptotes at x = ± 2,
so the function is continuous on:
-3 [ x < -2 and -2 < x < 2 and 2 < x [ 3. 2) is continuous for x m 0
and x² - 1 has vertical asymptotes at x = ± 1
the function is continuous on x m 0, x ! 1.

3)
continuous for all Real numbers, x ! 2
removeable jump at x = 2, y = 1/12.
4) -2 [ x [ 3 ensures that the expression
is always positive or zero so that the
square root exists.

2) Find values for c and/or c and d so that f(x) is continuous for all Real numbers.

1)
set them equal, solve for c = 5/2 or 2.5
2)

set them equal, solve for c = - 2.

3)

Now we solve the system of 2 equations
for c = -2 and d = -6.
4)

(c - 2)(c - 1) = 0 gives solution c = 1 or 2.

Back to Calculus I MathRoom Index

1) so limit = 7	2) so limit = 1

3) therefore, limit doesn't exist.

1) is restricted to -3 [ x [ 3 and x⁴ - 16 has vertical asymptotes at x = ± 2, so the function is continuous on: -3 [ x < -2 and -2 < x < 2 and 2 < x [ 3.	2) is continuous for x m 0 and x² - 1 has vertical asymptotes at x = ± 1 the function is continuous on x m 0, x ! 1.

3) continuous for all Real numbers, x ! 2 removeable jump at x = 2, y = 1/12.	4) -2 [ x [ 3 ensures that the expression is always positive or zero so that the square root exists.

1) set them equal, solve for c = 5/2 or 2.5	2) set them equal, solve for c = - 2.

3) Now we solve the system of 2 equations for c = -2 and d = -6.	4) (c - 2)(c - 1) = 0 gives solution c = 1 or 2.