Divergent limit + divergent limit = convergent limitIs it possible?

[gikiian]

Homework Statement

As a part of Method of Frobenius, I am encountered with the following problems:

Evaluate the following limits:

Q1. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}$

Q2. $\stackrel{limit}{_{x→0}}\frac{x-1}{x}$

Q3. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}$

In context of the above problems, I am having difficulty in verifying the following property of limits:

$\stackrel{limit}{_{x→a}}(f(x)+g(x))=\stackrel{limit}{_{x→a}}f(x)+ \stackrel{limit}{_{x→a}}g(x)$​

Homework Equations

N/A

The Attempt at a Solution

A1. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}=∞$ (i.e. the limit is divergent)

A2. $\stackrel{limit}{_{x→0}}\frac{x-1}{x}=-∞$ (i.e. the limit is divergent)

A3. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}=\frac{1-2x+x-1}{x}=\frac{-x}{x}=-1$ (i.e. the limit is convergent)I am just confused by the apparent fact that sum of two divergent limits can be a convergent limit. Even though this is apparent in my problem, I still want to make sure if I am not messing up anywhere.

 

[Ray Vickson]

Homework Statement

As a part of Method of Frobenius, I am encountered with the following problems:

Evaluate the following limits:

Q1. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}$

Q2. $\stackrel{limit}{_{x→0}}\frac{x-1}{x}$

Q3. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}$

In context of the above problems, I am having difficulty in verifying the following property of limits:

$\stackrel{limit}{_{x→a}}(f(x)+g(x))=\stackrel{limit}{_{x→a}}f(x)+ \stackrel{limit}{_{x→a}}g(x)$​

Homework Equations

N/A

The Attempt at a Solution

A1. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}=∞$ (i.e. the limit is divergent)

A2. $\stackrel{limit}{_{x→0}}\frac{x-1}{x}=-∞$ (i.e. the limit is divergent)

A3. $\stackrel{limit}{_{x→0}}\frac{1-2x}{x}+\frac{x-1}{x}=\frac{1-2x+x-1}{x}=\frac{-x}{x}=-1$ (i.e. the limit is convergent)


I am just confused by the apparent fact that sum of two divergent limits can be a convergent limit. Even though this is apparent in my problem, I still want to make sure if I am not messing up anywhere.

You cannot verify that "result" because it is just false, as your example shows. There is a somewhat similar true result, but it involves some extra hypotheses that your example fails to satisfy.

 

[shortydeb]

you're not summing two divergent limits. You're summing two functions whose individual limits at the same point are both divergent.

 

[LCKurtz]

In context of the above problems, I am having difficulty in verifying the following property of limits:

$\stackrel{limit}{_{x→a}}(f(x)+g(x))=\stackrel{limit}{_{x→a}}f(x)+ \stackrel{limit}{_{x→a}}g(x)$​

What the theorem states is that if both limits on the right side exist, then the limit on the left exists and equals the sum of the limits on the right. That's all.