Limits, continuity and L Hospital rule
Read a limit from three sides, spot discontinuities, and resolve 0/0 forms.
- Evaluate left-hand and right-hand limits and decide when a two-sided limit exists
- Classify removable, jump and infinite discontinuities from a formula
- Apply L Hospital rule only after verifying the form is 0/0 or infinity/infinity
- Combine limit laws (sum, product, squeeze) without guessing
An examiner writes lim(x->0) sin(x)/x on the board and asks for the answer without a calculator. The ones who understand that the ratio of two quantities going to zero at the same rate approaches the ratio of their instantaneous speeds write 1 in five seconds. That speed difference is what this lesson builds.
A limit answers one question: what value does f(x) approach as x approaches a, even if f(a) itself is undefined.
Two-sided limit exists iff:
- Left-hand limit lim(x->a-) f(x) exists.
- Right-hand limit lim(x->a+) f(x) exists.
- They are equal.
Continuity at a: the limit exists AND equals f(a).
| Discontinuity type | Behaviour | Example |
|---|---|---|
| Removable | Limit exists, f(a) differs or undefined | (x^2-1)/(x-1) at x=1 |
| Jump | Left and right limits finite but unequal | Sign function at 0 |
| Infinite | One or both sides shoot to infinity | 1/x at x=0 |
L Hospital rule: for form 0/0 or infinity/infinity, lim f/g = lim f'/g', provided the new limit exists. It does NOT apply to 0 x infinity or infinity - infinity without rewriting first.
Substitute the value mentally first. If the result is not 0/0 or infinity/infinity, L Hospital is the wrong tool and an examiner will award zero.
| Form | How to handle |
|---|---|
| 0/0 | L Hospital directly |
| infinity/infinity | L Hospital directly |
| 0 x infinity | Rewrite one factor as 1 over something, then get 0/0 |
| infinity - infinity | Combine into a single fraction, then get 0/0 |
| 1^infinity, 0^0, infinity^0 | Take log, resolve the log limit, then exponentiate |
- 1
Check the form
Substitute x = 0: numerator 1 - 1 = 0, denominator 0. Form is 0/0, so L Hospital is legal.
(1 - cos 0) / 0^2 = 0 / 0 - 2
Apply L Hospital once
Differentiate: sin x over 2x. Substitute again: still 0/0.
sin x / 2x -> 0/0 again - 3
Apply L Hospital a second time
Differentiate again: cos x over 2. Substitute: cos 0 / 2 = 1/2.
cos x / 2 -> 1/2
Solution
The limit is 1/2. One application of L Hospital is not always enough. Re-check the form after differentiating; if still indeterminate, apply again.
1. lim(x->0) sin(3x) / x 2. lim(x->0) (e^x - 1) / x 3. lim(x->2) (x^2 - 4) / (x - 2) 4. lim(x->infinity) (3x^2 + x) / (x^2 - 5)
Hint: 1 and 2 are 0/0: L Hospital once each. 3 factors. 4: compare highest powers, answer 3.
Applying L Hospital to a non-indeterminate form
lim(x->1) x^2 = 1 by direct substitution. Differentiating gives 2x = 2, wrong. Always substitute first.
Forgetting to re-check the form after differentiating
After each L Hospital application, substitute again. If no longer 0/0 or infinity/infinity, stop.
Confusing continuity with differentiability
Differentiable implies continuous; |x| is continuous at 0 yet not differentiable there.
Show lim(x->0) (tan x - x) / x^3 = 1/3.
Form 0/0. Differentiate: (sec^2 x - 1) / 3x^2 = tan^2 x / 3x^2. Still 0/0: differentiate again or use tan x ~ x near 0 to get x^2/(3x^2) = 1/3.
Source: Standard AKTU sessional
- Two-sided limit exists iff left limit = right limit, both finite.
- Continuity = limit exists AND equals f(a).
- L Hospital: only 0/0 or infinity/infinity; re-check form after each application.
- Differentiability implies continuity; the converse is false (|x| at 0).
