Limits, Continuity, and Discontinuities
Limits and Infinite Behavior
Concept of a limit: as x\to a (from either side), the values of the function f(x) approach a number A (which need not be the actual value f(a) at the point). The idea is captured by the notation \lim_{x\to a} f(x)=A\,.
One-sided limits: the statement above is really about approaching from both sides; sometimes the left-hand limit and right-hand limit can behave differently.
Infinite limits and vertical asymptotes:
If \lim_{x\to a} f(x)=+\infty (or -\infty), the graph shows unbounded growth near x=a and we say there is a vertical asymptote at x=a.
This behavior must occur from both sides in the sense that the two one-sided limits are both infinite (same sign) or diverge to infinity; the two-sided limit does not exist.
If the one-sided limits are different in sign or finite values (even if the graph shoots to infinity on one side and stays finite on the other), we still typically describe a vertical asymptote at that point because the function is unbounded near it.
Important takeaway about limits and values:
The value of the function at the limit point (i.e., f(a)) is irrelevant for the limit itself. Limits depend on behavior of f(x) for x\neq a near a, not on the value at a.
Squeeze Theorem
Setup: Suppose there are functions f(x)\le g(x)\le h(x) for all x (or for all x in a punctured neighborhood of) a, and
\lim{x\to a} f(x)=\lim{x\to a} h(x)=L\,.
Then, by the Squeeze Theorem,
\lim_{x\to a} g(x)=L\,.
The intuitive picture: f and h trap g between them as x approaches a, forcing g to have the same limit L.
Practical use: to determine limits of middle functions by sandwiching them between two known functions with the same limit.
Typical form and application of the Squeeze Theorem
Example: g(x)=x\sin\left(\frac{1}{x}\right) as x\to 0.
We have the bounding inequalities
For x>0: (-x\le x\sin(1/x)\le x).
For x<0: (-|x|\le x\sin(1/x)\le |x|) (in short, |x\sin(1/x)|\le |x|).
Therefore, - |x| \le x\sin(1/x) \le |x|\to 0 as x\to 0, so by the Squeeze Theorem
\lim_{x\to 0} x\sin(1/x)=0.
Example: Indeterminacy 0/0 and the need for nontrivial tricks (WTF moment):
Some limits are not approachable by straightforward substitution or simple algebraic manipulation (they are indeterminate forms, e.g., \frac{0}{0}).
One often uses known limits, substitutions, or trigonometric inequalities to resolve them.
Another powerful trick: substitution to reduce to a known limit.
Example: \lim_{x\to 1} \frac{\sin(\ln x)}{\ln x}
Let t=\ln x. As x\to 1, t\to 0.
Then the limit becomes \lim_{t\to 0} \frac{\sin t}{t}=1, which is the standard limit.
After computing in terms of t, revert the substitution to return to the original variable. Note that the limit value is a number; the intermediate variable is just a tool.
Important caveat about substitutions:
When you substitute a new variable, you must ensure the new limit in terms of the substituted variable corresponds to the original limit, and then reinterpret the result in terms of the original variable.
Practical approach to limits (summary):
1) Try direct substitution. If it yields a determinate number, that is the limit.
2) If substitution yields indeterminate form (e.g., 0/0), try algebraic manipulation (factoring, canceling common factors, etc.).
3) If algebraic manipulation fails or is not possible, consider other methods (trigonometric inequalities, squeeze, substitutions, l'Hôpital’s rule in more advanced contexts).
4) In many cases, transforming the limit to a known standard limit (like \lim_{t\to 0} \frac{\sin t}{t}=1) is the key.Note: The famous limit \lim_{x\to 0} \frac{\sin x}{x}=1 is a nontrivial limit that often serves as a building block for others and is typically memorized or derived via trigonometric inequalities and the squeeze theorem.
The limit that requires trigonometry and the squeeze theorem (detailed version)
A standard derivation uses a unit circle with angle \theta (take \theta in radians) and the triangle relations:
In the unit circle, the segment lengths satisfy inequalities such as
\sin \theta \le \theta \le \tan \theta\, for small positive \theta (and similarly for negative values by symmetry).
Divide the inequality by \sin \theta>0 (for small positive \theta):
1 \le \frac{\theta}{\sin \theta} \le \frac{1}{\cos \theta}\,.
Take reciprocals (reversing the inequality signs appropriately):
\cos \theta \le \frac{\sin \theta}{\theta} \le 1\,.
Let \theta\to 0: \cos \theta \to 1, hence by the Squeeze Theorem
\lim_{\theta\to 0} \frac{\sin \theta}{\theta}=1\,.
Replacing \theta by x gives the essential limit
\lim_{x\to 0} \frac{\sin x}{x}=1\,.
This limit is fundamental because many limits can be reduced to it via substitutions or small-angle approximations.
Continuity: definition and intuition
Definition: A function f is continuous at a point a if
the limit \lim_{x\to a} f(x) exists, and
this limit equals the function value at the point: \lim_{x\to a} f(x)=f(a)\,.
If the limit exists but is not equal to f(a), the function has a discontinuity at a; the limit itself is well-defined, but the function value is not what the limit would suggest.
Visual intuition with a few scenarios:
If the left-hand and right-hand limits both exist and equal the same value L and f(a)=L, then f is continuous at a.
If the left-hand and right-hand limits exist and equal some value L\neq f(a), we have a removable discontinuity (can be "fixed" by redefining f(a)=L).
If the left-hand and right-hand limits exist but are not equal (or if one-sided limits do not exist), the two-sided limit does not exist; this is a discontinuity of a different kind (e.g., a jump discontinuity or infinite discontinuity, depending on the behavior).
Two-sided limit and continuity link: continuity at a requires both the limit to exist and to equal the function value at a.
A point to remember: continuity is a local property; a function can be continuous at a point without being continuous elsewhere.
Types of discontinuities
Removable discontinuity:
The limit exists (finite) but f(a)\neq \lim_{x\to a} f(x).
Can be "removed" by redefining f(a) to the limit value.
Jump discontinuity:
The left-hand limit and the right-hand limit exist but are not equal:
\lim{x\to a^-} f(x) \neq \lim{x\to a^+} f(x).
The function value at a may be anything; the discontinuity cannot be fixed by a single redefining of the value.
Infinite (essential) discontinuity / vertical asymptote:
The limit from one or both sides diverges to +\infty or -\infty:
Example: vertical asymptote at x=a when \lim_{x\to a} f(x)=\pm\infty.
Example recap from the notes:
A graph with a vertical asymptote at x=a shows one-sided limits going to ±∞ on either side.
A function like \frac{\sin(1/x)}{x} near x=0 can fail to have a limit if the oscillation becomes unbounded; such behavior is a kind of non-removable discontinuity.
Discontinuities and continuity on intervals
A function is continuous on an interval if it is continuous at every point in that interval.
Polynomial functions are continuous everywhere on the real line (a standard example of continuous functions).
The notion of continuity on intervals underpins many foundational results in analysis and calculus.
Summary takeaways:
Limits describe local behavior near a point, independent of the function's value there.
Continuity ties the limit to the function value at the point.
Most common discontinuities are removable, jump, and infinite; infinite is essentially tied to vertical asymptotes.
The Squeeze Theorem provides a robust tool for proving limits by trapping a function between two others with the same limit.
Known limits (like \lim_{x\to 0}\frac{\sin x}{x}=1) and algebraic/trigonometric tricks are essential to solving limits that resist direct substitution.
Quick references and practical tips
When evaluating limits in practice:
Start with substitution. If it works, you’re done.
If substitution yields an indeterminate form like 0/0, search for algebraic simplifications.
If algebra is unproductive, look for a squeeze argument or a trigonometric substitution/inequality.
Consider known standard limits and substitutions to reduce to these limits.
Remember the key limit for trigonometry: \lim_{x\to 0} \frac{\sin x}{x}=1, and how to justify it via the Squeeze Theorem.
The value of the function at the limit point does not affect the limit; continuity depends on matching the two.
For a function to be continuous at a, both the limit exists and equals the function value: \lim_{x\to a} f(x)=f(a).
End of notes
We can continue with more continuity topics and examples in the next session.