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Basic Limit Properties

April 8th, 2009 |

Author: Merrin

Although we will not be using the formal definition of limits which can be stated in terms of what is called a delta-epsilon proof, we will use some basic limit properties which have already been proven to do our basic manipulations.

Theorem 1. Basic limit laws. Given that

$\mathop{\lim }\limits_{x\to a} f(x)=M\quad \mathop{\lim }\limits_{x\to a} g(x)=N$

Then the following facts are true.

$\mathop{\lim }\limits_{x\to a} [f(x)\pm g(x)]=\mathop{\lim }\limits_{x\to a} f(x)\pm \mathop{\lim }\limits_{x\to b} g(x)=M\pm N$
$\mathop{\lim }\limits_{x\to a} cf(x)=c\mathop{\lim }\limits_{x\to a} f(x)=cM$
$\mathop{\lim }\limits_{x\to a} [f(x)g(x)]=\mathop{\lim }\limits_{x\to a} f(x)\mathop{\lim }\limits_{x\to a} g(x)=MN$
$\mathop{\lim }\limits_{x\to a} \cfrac{f(x)}{g(x)} =\cfrac{\mathop{\lim }\limits_{x\to a} f(x)}{\mathop{\lim }\limits_{x\to a} g(x)} =\cfrac{M}{N} ,\quad \mathop{\lim }\limits_{x\to a} g(x)\ne 0$

Example 1. Give an example when limits cannot be combined.

Solution 1. Well the theorem for the basic limit laws requires that the separate limits exist.

$\mathop{\lim }\limits_{x\to 0} \left(\cfrac{1}{x^{2} } -\cfrac{1}{x^{2} } \right)=0\quad \, \text{but}\quad \, \mathop{\lim }\limits_{x\to 0} \cfrac{1}{x^{2} } -\mathop{\lim }\limits_{x\to 0} \cfrac{1}{x^{2} } =DNE$

The sum of two infinite limits taken separately cannot be combined on the right. I don’t necessarily agree with this but it is the law! Taking a limit is a sort of processing that can’t be undone.

Example 2. The basic limit properties can be useful in simplifying calculations. Evaluate

$\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x\tan x}{x^{2} }$

Given that

$\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} =1$

Solution 2. Using the law for the product of limits we can write

${\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x\tan x}{x^{2} } } ={\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} \cdot \mathop{\lim }\limits_{x\to 0} \cfrac{\tan x}{x} }$

$= {\mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} \mathop{\lim }\limits_{x\to 0} \cfrac{\sin x}{x} \mathop{\lim }\limits_{x\to 0} \cfrac{1}{\cos x} }$

$={1\cdot 1\cdot 1=1}$

We have used the given limit, the fact that secant is continuous at x = 0, and the product rule for limits.

Example 3. Evaluate the following limit

$\mathop{\lim }\limits_{x\to 2} \cfrac{x-\ln x}{\sin x}$

Solution 3. Using all the limit basic laws successively we have

$\mathop{\lim }\limits_{x\to 2} \cfrac{x-\ln x}{\sin x} =\cfrac{\mathop{\lim }\limits_{x\to 2} x-\mathop{\lim }\limits_{x\to 2} \ln x}{\mathop{\lim }\limits_{x\to 2} \sin x} =\cfrac{2-\ln 2}{\sin 2}$

The function is continuous, but the limit laws also apply.

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Basic Limit Properties

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