Properties of Limits and Calculation Techniques

In the introduction to limits, we saw that

$$\lim_{x\to a}f(x)=L$$

means that the values of $f(x)$ approach $L$ as $x$ approaches $a$.

Now we will develop tools that allow us to calculate limits systematically.

The main ideas in this topic are:

• algebraic properties of limits;
• direct substitution;
• factoring;
• simplifying expressions;
• rationalization;
• one-sided limits;
• identifying indeterminate forms.

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1. Sum Property

If the limits

$$\lim_{x\to a}f(x)=L$$

and

$$\lim_{x\to a}g(x)=M,$$

exist, then

$$\boxed{ \lim_{x\to a}[f(x)+g(x)]=L+M }$$

Example

Calculate

$$\lim_{x\to2}(x^2+3x).$$

We can separate the limit:

$$\lim_{x\to2}x^2+ \lim_{x\to2}3x.$$

Then,

$$2^2+3(2)=4+6.$$

Therefore,

$$\boxed{ \lim_{x\to2}(x^2+3x)=10 }$$

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2. Difference Property

Similarly,

$$\boxed{ \lim_{x\to a}[f(x)-g(x)] = L-M }$$

Example

$$\lim_{x\to3}(x^2-2x).$$

We have

$$3^2-2(3)=9-6.$$

Therefore,

$$\boxed{ \lim_{x\to3}(x^2-2x)=3 }$$

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3. Product Property

If

$$\lim_{x\to a}f(x)=L$$

and

$$\lim_{x\to a}g(x)=M,$$

then

$$\boxed{ \lim_{x\to a}[f(x)g(x)]=LM }$$

Example

$$\lim_{x\to2}(x+1)(x^2+3).$$

We can substitute directly:

$$(2+1)(2^2+3).$$ $$=3\cdot7.$$

Therefore,

$$\boxed{21}.$$

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4. Constant Property

If $c$ is a constant, then

$$\boxed{ \lim_{x\to a}c=c }$$

Furthermore,

$$\boxed{ \lim_{x\to a}[cf(x)] = c\lim_{x\to a}f(x) }$$

Example

If

$$\lim_{x\to2}f(x)=5,$$

then

$$\lim_{x\to2}3f(x) = 3\cdot5 = 15.$$

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5. Quotient Property

If

$$\lim_{x\to a}f(x)=L$$

and

$$\lim_{x\to a}g(x)=M,$$

with $M\neq0$, then

$$\boxed{ \lim_{x\to a} \frac{f(x)}{g(x)} = \frac{L}{M} }$$

Example

Calculate

$$\lim_{x\to2} \frac{x^2+1}{x+3}.$$

Since the denominator does not equal zero at $x=2$, we can substitute directly:

$$\frac{2^2+1}{2+3} = \frac{5}{5}.$$

Therefore,

$$\boxed{ \lim_{x\to2} \frac{x^2+1}{x+3} =1 }$$

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6. Limits of Powers

If

$$\lim_{x\to a}f(x)=L,$$

then, for a natural number $n$,

$$\boxed{ \lim_{x\to a}[f(x)]^n=L^n }$$

Example

$$\lim_{x\to2}(x+1)^3.$$

Substituting:

$$(2+1)^3=3^3.$$

Therefore,

$$\boxed{27}.$$

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7. Limits of Roots

When the expression is defined in the domain under consideration, we can use

$$\boxed{ \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{L} }$$

Example

$$\lim_{x\to4}\sqrt{x+5}.$$

Substituting $x=4$:

$$\sqrt{4+5} = \sqrt9 = 3.$$

Therefore,

$$\boxed{ \lim_{x\to4}\sqrt{x+5}=3 }$$

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8. Direct Substitution

The properties above allow us to solve many limits simply by substituting $x=a$.

For example:

$$\lim_{x\to3} \frac{x^2+2x-1}{x+1}.$$

Substituting $x=3$:

$$\frac{3^2+2(3)-1}{3+1}.$$ $$= \frac{9+6-1}{4}.$$ $$=\frac{14}{4}.$$

Therefore,

$$\boxed{ \lim_{x\to3} \frac{x^2+2x-1}{x+1} = \frac72 }$$

Attention

Direct substitution should be the first attempt.

However, it does not always immediately give us the limit.

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9. When $\frac00$ Appears

Consider:

$$\lim_{x\to2} \frac{x^2-4}{x-2}.$$

Substituting directly:

$$\frac{2^2-4}{2-2} = \frac00.$$

In this case, we cannot conclude that the limit is zero or that it does not exist.

The expression

$$\boxed{\frac00}$$

is an indeterminate form.

We need to transform the expression before calculating the limit.

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10. Factoring Technique

In the previous example, we have

$$x^2-4.$$

Using the difference of squares:

$$x^2-4=(x-2)(x+2).$$

Thus,

$$\frac{x^2-4}{x-2} = \frac{(x-2)(x+2)}{x-2}.$$

For $x\neq2$:

$$\frac{x^2-4}{x-2}=x+2.$$

Therefore,

$$\lim_{x\to2} \frac{x^2-4}{x-2} = \lim_{x\to2}(x+2).$$

Now we can substitute:

$$2+2=4.$$

Therefore,

$$\boxed{ \lim_{x\to2} \frac{x^2-4}{x-2} =4 }$$

Important Idea

Even if the original expression is not defined at $x=2$, the limit may still exist.

What matters is the behavior of the function near $2$.

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11. Another Factoring Example

Calculate

$$\lim_{x\to1} \frac{x^2+x-2}{x-1}.$$

Substituting directly:

$$\frac{1+1-2}{1-1} = \frac00.$$

We have an indeterminate form.

Factoring the numerator:

$$x^2+x-2=(x-1)(x+2).$$

Then,

$$\frac{x^2+x-2}{x-1} = x+2, \qquad x\neq1.$$

Therefore,

$$\lim_{x\to1} \frac{x^2+x-2}{x-1} = 1+2.$$

Thus,

$$\boxed{3}.$$

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12. Rationalization Technique

Another very common situation occurs when we have square roots.

Consider:

$$\lim_{x\to4} \frac{\sqrt{x}-2}{x-4}.$$

Direct substitution produces

$$\frac{\sqrt4-2}{4-4} = \frac00.$$

We again have an indeterminate form.

In this case, we can use the conjugate.

The conjugate of

$$\sqrt{x}-2$$

is

$$\sqrt{x}+2.$$

We multiply the numerator and denominator by the conjugate:

$$\frac{\sqrt{x}-2}{x-4} \cdot \frac{\sqrt{x}+2}{\sqrt{x}+2}.$$

The numerator becomes a difference of squares:

$$(\sqrt{x}-2)(\sqrt{x}+2)=x-4.$$

Thus,

$$\frac{(x-4)} {(x-4)(\sqrt{x}+2)}.$$

For $x\neq4$:

$$\frac{1}{\sqrt{x}+2}.$$

Therefore,

$$\lim_{x\to4} \frac{\sqrt{x}-2}{x-4} = \lim_{x\to4} \frac{1}{\sqrt{x}+2}.$$

Substituting:

$$\frac{1}{2+2} = \frac14.$$

Therefore,

$$\boxed{ \lim_{x\to4} \frac{\sqrt{x}-2}{x-4} = \frac14 }$$

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13. One-Sided Limits

We have already seen that we can approach a point from the left or from the right.

The left-hand limit is

$$\lim_{x\to a^-}f(x).$$

The right-hand limit is

$$\lim_{x\to a^+}f(x).$$

For the two-sided limit to exist, we need

$$\boxed{ \lim_{x\to a^-}f(x) = \lim_{x\to a^+}f(x) }$$

Example

Consider a function such that

$$\lim_{x\to2^-}f(x)=4$$

and

$$\lim_{x\to2^+}f(x)=4.$$

Then,

$$\boxed{ \lim_{x\to2}f(x)=4 }$$

On the other hand, if

$$\lim_{x\to2^-}f(x)=4$$

and

$$\lim_{x\to2^+}f(x)=7,$$

then

$$\boxed{ \lim_{x\to2}f(x) \text{ does not exist} }$$

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14. A Strategy for Calculating Limits

When encountering a limit, we can follow a sequence of steps.

Step 1 — Try Direct Substitution

Calculate

$$f(a).$$

If the result is a well-defined real number, the limit is generally determined.

Step 2 — Check Whether an Indeterminate Form Appears

The most common forms are

$$\frac00$$

and

$$\frac{\infty}{\infty}.$$

These expressions are not results of limits. They indicate that we need to continue working with the expression.

Step 3 — Look for a Simplification

Depending on the expression, we can use:

• factoring;
• notable products;
• rationalization;
• reducing to a common denominator;
• trigonometric identities.

Step 4 — Calculate the Limit Again

After simplifying, try direct substitution again.

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15. Complete Example

Calculate

$$\lim_{x\to3} \frac{x^2-9}{x-3}.$$

Direct Substitution

$$\frac{3^2-9}{3-3} = \frac00.$$

We have an indeterminate form.

Factoring

$$x^2-9=(x-3)(x+3).$$

Therefore,

$$\frac{x^2-9}{x-3} = x+3, \qquad x\neq3.$$

Thus,

$$\lim_{x\to3} \frac{x^2-9}{x-3} = \lim_{x\to3}(x+3).$$

Finally,

$$3+3=6.$$

Therefore,

$$\boxed{ \lim_{x\to3} \frac{x^2-9}{x-3}=6 }$$

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16. Summary of the Main Properties

If

$$\lim_{x\to a}f(x)=L$$

and

$$\lim_{x\to a}g(x)=M,$$

then:

Sum

$$\boxed{ \lim_{x\to a}[f(x)+g(x)]=L+M }$$

Difference

$$\boxed{ \lim_{x\to a}[f(x)-g(x)]=L-M }$$

Product

$$\boxed{ \lim_{x\to a}[f(x)g(x)]=LM }$$

Quotient

If $M\neq0$,

$$\boxed{ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{L}{M} }$$

Constant

$$\boxed{ \lim_{x\to a}c=c }$$

Power

$$\boxed{ \lim_{x\to a}[f(x)]^n=L^n }$$

Root

When defined,

$$\boxed{ \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{L} }$$

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17. Main Techniques

Situation	Strategy
Substitution gives a number	Direct substitution
$\frac00$ appears in polynomials	Factoring
Roots appear	Rationalization
Piecewise-defined function	One-sided limits
Trigonometric expressions	Trigonometric identities
Limits at infinity	Comparison of dominant terms

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18. What Comes Next?

With these properties and techniques, we can now move on to some of the most important limits in Calculus I.

One of the next fundamental results is:

$$\boxed{ \lim_{x\to0}\frac{\operatorname{sen}x}{x}=1 }$$

This limit will later be used in the derivation of trigonometric functions.

From it, we can also study limits involving

$$\frac{1-\cos x}{x}, \qquad \frac{\tan x}{x},$$

as well as exponential and logarithmic limits.