Calculus Solver How to find limits
Try direct substitution first. If you get a number, that is the limit and you are done. If you get 0/0, the expression can be simplified by factoring, by multiplying by a conjugate, or by dividing through by the highest power, and then substitution works. If you get a non-zero number over 0, the limit is infinite or does not exist.
What a limit is actually asking
The limit as x approaches a is the value the function heads toward as x gets close to a, from both sides. It is not the value at a. A function can have a hole at x = 2 and still have a perfectly good limit there, which is exactly the situation the diagram shows and the reason limits exist as a concept at all.
This is why substitution failing does not mean the limit fails. It means the formula is undefined at that one point, and the job is to rewrite the formula into one that is not.
1. Substitution, always first
Put the value in. For lim as x approaches 3 of (x² + 1), substitution gives 10, and that is the answer. Any function that is continuous at the point gives up its limit this way, which covers polynomials everywhere, and rational, root, exponential and trigonometric functions wherever they are defined.
Most limit problems on a test are designed so that substitution fails, but you still start here, because the form you get tells you which method comes next.
2. Factor and cancel, when you get 0/0
0/0 means numerator and denominator share a factor that is zero at your point. Factor both, cancel it, then substitute into what is left.
For lim as x approaches 2 of (x² − 4)/(x − 2): substitution gives 0/0. Factor the top into (x − 2)(x + 2), cancel the (x − 2), and you are left with x + 2. Substituting now gives 4.
The cancelled factor is the hole. The original function has no value at x = 2, but it approaches 4, and 4 is the limit.
3. Multiply by the conjugate, when there is a root
When 0/0 comes with a square root, factoring will not reach it. Multiply top and bottom by the conjugate, which is the same expression with the sign between the terms flipped.
For lim as x approaches 0 of (√(x + 9) − 3)/x, multiply through by (√(x + 9) + 3). The numerator becomes (x + 9) − 9 = x, so the whole thing collapses to 1/(√(x + 9) + 3). Substituting gives 1/6.
4. Divide by the highest power, for limits at infinity
When x approaches infinity and you get ∞/∞, divide every term by the highest power of x in the denominator. Terms of the form constant over a power of x then go to zero and the answer falls out.
For lim as x approaches ∞ of (3x² + 5x)/(2x² − 1), divide everything by x²: (3 + 5/x)/(2 − 1/x²). The 5/x and 1/x² vanish, leaving 3/2.
The shortcut worth memorising is that only the leading degrees matter. Equal degrees give the ratio of the leading coefficients, a bigger degree on top gives infinity, and a bigger degree on the bottom gives zero.
When the limit does not exist
If the one-sided limits disagree, there is no limit. That is what happens at a jump, and at lim as x approaches 0 of |x|/x, where the left side heads to −1 and the right side to 1.
If substitution gives a non-zero number over zero, the function blows up. Report it as +∞ or −∞ if both sides agree, and as not existing if they run in opposite directions, which is the usual case at a vertical asymptote.
What the form you get is telling you
| Substitution gives | Meaning | What to do |
|---|---|---|
| A number | The function is continuous there | That number is the limit |
| 0/0 | Indeterminate, shared factor | Factor, or use a conjugate |
| ∞/∞ | Indeterminate | Divide by the highest power |
| k/0, k ≠ 0 | Vertical asymptote | ±∞, or does not exist |
| 0/k, k ≠ 0 | Not indeterminate | The limit is 0 |
Questions
Why does 0/0 not just mean the limit is 1 or 0?
Because 0/0 is not a value, it is a report that the algebra cancelled itself out. Different functions that both give 0/0 have different limits, so the form tells you nothing except that the expression needs rewriting.
When can I use L'Hôpital's rule?
Only on 0/0 or ∞/∞. Differentiate the top and the bottom separately, not as a quotient, and try the limit again. It is quicker than factoring on hard expressions, but applying it to a form that is not indeterminate produces a wrong answer, so check the form first.
Does the function have to be defined at the point?
No, and that is the point of a limit. It describes where the function is heading near a, and deliberately ignores what happens at a.
What is a one-sided limit?
The value approached from one direction only, written with a superscript minus for from the left and plus for from the right. The two-sided limit exists only when both agree, which is the standard way to prove a limit does not exist at a jump.
More calculus reference
Stuck on the actual problem
Photograph it and get the worked solution, step by step, with the rule that justified each line named.
