*Notes to a video lecture on http://www.unizor.com*

__Function Limit - Exercise__

(

(

*x→−∞*)Try to do these exercises yourself.

All function limits below are supposed to be calculated as argument

**tends to negative infinity, that is infinitely decreasing without bounds, eventually getting less than any real number fixed beforehand and staying below that number ever since.**

*x*In other words, we say that

**as**

*f(x)→L***, if**

*x→−∞***∀**positive

**(however small)**

*ε***∃**

*A*: (*x ≤ A*)⇒|*f(x)−L*| ≤*ε*In some cases it's useful to use the following

*Theorem*

If

**as**

*f(x)→L***,**

*x→−∞*then

**as**

*f(−x)→L*

*x→+∞*and

*converse*(former follows from latter).

*Proof*

Actually, this theorem is following from a more general theorem about limit of a compounded function.

Indeed, set

**and consider limit of function**

*g(x)=−x***as**

*f(g(x))***.**

*x→+∞*However, it might be useful to prove this theorem directly.

Given that

**∀**positive

**(however small)**

*ε***∃**

*A*: (*x*≤*A*) ⇒ |*f(x)−L*| ≤*ε*Choose positive

**(however small).**

*ε*Let

**.**

*B = −A*Let

**.**

*x' = −x*For any

*x ≥ B*it is true that

**.**

*−x ≤ −B*Therefore,

**.**

*x' ≤ A*Therefore,

**|**

*f(x')−L*| ≤*ε*Since

**,**

*f(x') = f(−x)*it's true that

**|**

*f(−x)−L*| ≤*ε*So, we have proven that

**∀**

(

*ε*>*0*∃*B*:(

*x*≥*B*) ⇒ |*f(−x)−L*| ≤*ε*That is,

*x→+∞*⇒*f(−x)→L*End of proof.

Find the limits of the following functions as their argument

**infinitely decreasing.**

*x*1.

*(2x²+3x+4)/(3x²+4x+5)**Answer*:

*2/3*2.

*(2x³+3x+4)/(3x²+4x+5)**Answer*: Function will br infinitely decreasing or, non-rigorously, its limit is

*−∞*3.

*(2x²+3x+4)/(3x³+4x+5)**Answer*:

*0*4.

*x·sin(1/x)**Answer*:

*1*5.

*x·3*^{x}*Hint*: Switch to

**and the results of corresponding exercise from a previous topic.**

*x→+∞**Answer*:

*0*6.

*x*^{2}·3^{x}*Hint*: Switch to

**and the results of corresponding exercise from a previous topic.**

*x→+∞**Answer*:

*0*
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