Theory Recap - What Is Continuity?
A function is continuous at a point if and only if all three of the following conditions hold:
: 
exists and is a real number.: 
If any one of these conditions fails, the function is discontinuous at 
. A function continuous at every point in its domain is called a continuous function.
Many students check only that the limit exists, forgetting to verify that it also equals . Always check all three conditions — especially for piecewise functions where the formula changes at the boundary point.
Continuity Practice Problems And Solutions
Find the point(s) of discontinuity of the function:
Step 1 — Factor the denominator to find where it equals zero:
The function is undefined at 
and 
.
Step 2 — Factor the numerator to check whether either singularity cancels:
Step 3 — Classify each discontinuity.
At : the factor 
cancels, so 
exists, but 
is undefined. This is a removable discontinuity.
At : the denominator remains zero while the numerator equals 
. The function diverges — this is an infinite (essential) discontinuity.
∴ Discontinuities at (removable) and (infinite). The function is continuous everywhere else.
Consider the piecewise function defined for 
:
if 
if 
if 
Determine the values of a and b for which 
is continuous for all 
.
Continuity can only break at the boundary points and 
.
Step 1 — Continuity at :
For continuity: 
Step 2 — Continuity at :
For continuity: 
Step 3 — Solve the system: subtract (1) from (2):
∴ 
and 
Given the function:
Determine the value of a for which is continuous at 
.
For continuity at we need 
.
Use the standard result 
and multiply and divide by 3:
∴ 
Given the function:
Determine whether is continuous on 
.
For 
, the function is defined and continuous since 
only at .
At , direct substitution gives 
. Apply L'Hôpital's rule:
Since 
, all three continuity conditions are satisfied.
∴ is continuous on .
Given the function:
Determine a and b so that is continuous for all values of .
Step 1 — Continuity at :
For continuity:
Step 2 — Continuity at 
:
For continuity: 
∴ and 
Given the function:
Determine the value of a for which is continuous.
For continuity at we need 
. Factor the numerator:
∴ 
Find the value of k, if any exists, for the following function to be continuous at :
Evaluate the one-sided limits. Since 
for 
and 
for
The left-hand and right-hand limits are not equal, so the two-sided limit does not exist. This is a jump discontinuity. No value of 
can make the function continuous at .
∴ No value of makes continuous at . The function has a jump discontinuity there.
Given the function:
Determine the values of a and b in order to create a continuous function.
Step 1 — Continuity from the left at :
This holds for any value of , since the 
term vanishes at .
Step 2 — Continuity from the right at :
For continuity: 
∴ 
and can be any real number.
Determine the values of a and b in order to create a continuous function:
if 
if if 
Step 1 — Continuity at 
:
For continuity: 
Step 2 — Continuity at :
For continuity: 
Step 3 — Solve: substitute 
into (2):
∴ 
and 
Key Techniques Summary
Factoring and cancelling: the fastest way to resolve forms and identify removable discontinuities (Exercises 1, 6).
One-sided limits: always evaluate left and right limits separately at piecewise boundaries. If they differ, the overall limit does not exist (Exercise 7).
Standard trigonometric limit: . Rewrite the expression to match this form (Exercise 3).
L'Hôpital's rule: differentiate numerator and denominator separately when direct substitution gives or 
(Exercise 4).
Simultaneous equations: apply continuity at two boundary points to generate two equations, then solve for the two unknowns (Exercises 2, 5, 9).
For piecewise functions with two unknowns, you always need two boundary conditions. Students who find one equation and stop will score only partial marks. Check continuity at every boundary point.
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