Limit And Continuity Problems With Solution

Working Through Limits and Continuity by Hand

Most students treat limits and continuity like two separate topics. They're not. A limit tells you what a function is approaching. Continuity tells you whether the function actually lands on that value. The moment you see them as distinct, you'll miss whole classes of problems.

Common Limit And Continuity Problems With Solution Approaches

I spent years grading calc exams, and the pattern never changes. People can plug into L'Hôpital's rule without thinking, then get wrecked the second a piecewise function appears. Here's what actually works in practice. When you're given a piecewise function, stop and look at the boundary point first. Don't rush into simplifying. I once spent twenty minutes simplifying a rational expression, only to realize the function was undefined on one side entirely. That alone makes it discontinuous. Take a function like f(x) = (x² - 4)/(x - 2) when x 2, and f(2) = k. The limit as x approaches 2 exists and equals 4. But the function is only continuous if you define k = 4. Set k to anything else and you have a removable discontinuity. Students routinely pick k = 0 or k = 2 because they're rushing. It happens every semester.

When L'Hôpital's Rule Actually Fails

L'Hôpital's rule requires you to verify the indeterminate form first. Not always. This trips people up. Try it on lim(x0) x·sin(1/x). You might be tempted to rewrite it as sin(1/x)/(1/x), which gives sin(1/x)·x. That's still indeterminate-looking, but L'Hôpital's doesn't apply here because the derivative of sin(1/x) doesn't settle into a recognizable form. The actual answer is 0, found by the squeeze theorem instead. A jump discontinuity means the left-hand limit and right-hand limit both exist but aren't equal. A removable discontinuity means the two-sided limit exists but doesn't match the function's actual value at that point, or the function isn't defined there. For step functions, ceiling, floor — those create jump discontinuities everywhere integers appear. Don't try to compute derivatives there. Derivatives don't exist at jump discontinuities. The function isn't even continuous, so differentiability is out of the question.

A Problem I Keep Seeing Go Wrong

Here's a specific edge case. Consider f(x) = |x - 3|/(x - 3). What happens at x = 3? Left-hand limit: as x approaches 3 from below, the numerator is 3 - x and the denominator is x - 3, giving -1. Right-hand limit: as x approaches 3 from above, the numerator is x - 3 and the denominator is x - 3, giving 1.

Left right. Jump discontinuity. The limit doesn't exist. Simple, but students keep writing lim = 0 because they're averaging the two sides. That's not how limits work. There's no averaging step in the definition.

Continuity on Closed Intervals for the Intermediate Value Theorem

If you need to prove a root exists between a and b, the IVT requires continuity on the entire closed interval [a, b]. Not just at endpoints. I've seen solutions that check f(a) and f(b) signs but skip verifying continuity between them. That's incomplete. If there's a vertical asymptote somewhere inside, the theorem doesn't apply and your conclusion is wrong. Most students avoid epsilon-delta proofs entirely. Fair. They're tedious. But understanding the structure helps with rigorous limit problems. The basic pattern: given epsilon > 0, find delta > 0 such that |x - c| < delta implies |f(x) - L|

epsilon. For polynomial limits, you can usually pick delta = min(1, epsilon/(some bound)). For rational functions, factor and cancel first, then work with the simplified expression. Don't try delta-epsilon on everything. It's overkill for introductory problems.

Horizontal Asymptotes and End Behavior

Lim(x) of a rational function depends on degrees. Same degree: ratio of leading coefficients. Numerator higher degree: limit is ±. Denominator higher degree: limit is 0. Memorize this. It comes up constantly and saves time you'd otherwise waste doing synthetic division on infinite intervals. Just a note for anyone curious about the boundaries of this material. The Weierstrass function is continuous everywhere but differentiable nowhere. It's not relevant for a standard calc course, but it shows that continuity is genuinely a weaker condition than differentiability. You'll encounter this in advanced analysis. For limits: check if the expression is defined at the point. If not, try algebraic simplification. If it's a fraction approaching zero over zero, consider L'Hôpital's or conjugate multiplication. For radicals, conjugates usually help. For piecewise, compute one-sided limits separately.

For continuity at a point: three conditions must hold. The limit exists. The function is defined there. The limit equals the function value. Miss any one and it's discontinuous. That's the framework. Most problems fold into it without requiring anything deeper than careful algebra and attention to domain restrictions.