Calculus/Limits and their Properties/Evaluating Limits Analytically

Properties of Limits

Some Basic Limits

Let a and b be real numbers, and let n be a positive integer.

1. $\underset{x\to a}{\lim }b=b$
2. $\underset{x\to a}{\lim }x=a$
3. $\underset{x\to a}{\lim }{x}^n=a^n$

Example 1 | Evaluating Basic Limits

Properties of Limits

Let $\underset{x\to a}{\lim }f(x)=L_1$ and $\underset{x\to a}{\lim }g(x)=L_2$

1. $\underset{x\to a}{\lim }[f(x)+g(x)]=L_1+L_2$
2. $\underset{x\to a}{\lim }[f(x)\cdot g(x)]=L_1\cdot L_2$
3. $\underset{x\to a}{\lim }\frac{f(x)}{g(x)}=\frac{L_1}{L_2}$ if L₂ ≠ 0

Limits of Polynomial and Rational Functions

1. $\underset{x\to a}{\lim }{x}^n=a^n$ if n > 0

Example 2 | The Limit of a Rational Function

Find $\underset{x\to 2}{\lim }\frac{x^2+3x-10}{x-2}$.

Solution

Direct substitution gives the indeterminate form 0/0. The numerator can be seperated into the product of the two binomials (x + 5) and (x - 2).

So the limit is equivalent to $\underset{x\to 2}{\lim }\frac{(x+5)(x-2)}{(x-2)}$

From here, we can simply divide (x - 2) out of the fraction. We do not have to worry about (x - 2) being equal to 0, since in the context of this limit, the expression can be treated as if x will never equal 2.

This gives us $\underset{x\to 2}{\lim }(x+5)$. The expression inside the limit is now linear, so the limit can be found by direct substitution. This obtains 2 + 5 = 7.

We then can say that $\underset{x\to 2}{\lim }\frac{x^2+3x+10}{x-2}=7$.

The Limit of a Radical Function

1. $\underset{x\to a}{\lim }\sqrt[n]{x}=\sqrt[n]{a}$ if n > 0, and if n is even, then x > 0

The Limit of a Composite Function

Limits of Trigonometric Functions

The Squeeze Theorem

Definition

Two Special Trigonometric Limits

Definition

1. $\underset{x\to 0}{\lim }\frac{\sin x}{x}=1$
2. $\underset{x\to 0}{\lim }\frac{1-\cos x}{x}=0$

Example 3 | A Limit Involving a Trigonometric Function

Find the limit: $\underset{x\to 0}{\lim }\frac{\tan x}{x}$.

Solution

Direct substitution gives the indeterminate form 0/0. You can still solve this problem, however: write tan x as (sin x)/(cos x).

$\underset{x\to 0}{\lim }\frac{\tan x}{x}=\underset{x\to 0}{\lim }\left(\frac{\sin x}{x}\right)\left(\frac{1}{\cos x}\right)$.

Because

$\underset{x\to 0}{\lim }\frac{\sin x}{x}$ and $\underset{x\to 0}{\lim }\frac{1}{\cos x}$

so

$\underset{x\to 0}{\lim }\frac{\tan x}{x}$	=	$\left(\underset{x\to 0}{\lim }\frac{\sin x}{x}\right)\left(\underset{x\to 0}{\lim }\frac{1}{\cos x}\right)$
	=	(1)(1)
	=	1.