Limits and Continuity

limits

Limits

 

The curve y=a^{x} approaches the x-axis when x approaches very large negative numbers, but never touches it. That is x \rightarrow -\infty, a^{x} \rightarrow 0 .

We say that the limit of a^{x} as x approaches -\infty is 0. In symbols, we write:

\lim _{ x\rightarrow -\infty  }{ { a }^{ x } } =0

 

Examples

 

1. Find \lim _{ x\rightarrow 0 }{ \cfrac { { x }^{ 2 }+5x }{ x } }

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Substituting x=0 into the function \cfrac{0}{0}, which is undefined. Factorising and cancelling help us find the limit.

\lim _{ x\rightarrow 0 }{ \cfrac { { x }^{ 2 }+5x }{ x } } \\ =\lim _{ x\rightarrow 0 }{ \cfrac { x(x+5) }{ x } } \\ =\lim _{ x\rightarrow 0 }{ (x+5) } \\ =5

 

2. Find \lim _{ x\rightarrow 2 }{ \cfrac { x-2 }{ { x }^{ 2 }-4 } }

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Substituting x=2 into the function gives \cfrac{0}{0}, which is undefined.

\lim _{ x\rightarrow 2 }{ \cfrac { x-2 }{ { x }^{ 2 }-4 } } \\ =\lim _{ x\rightarrow 2 }{ \cfrac { x-2 }{ (x+2)(x-2) } } \\ =\lim _{ x\rightarrow 2 }{ \cfrac { 1 }{ x+2 } } \\ =\cfrac { 1 }{ 4 }

 

 

 

Continuity

 

Many functions are continuous, which means they have a smooth, unbroken curve (or line). However, several discontinuous functions have gaps in their graphs. If a curve is discontinuous at a certain point, we can use limits to find the value that the curve approaches at that point.

 

Example

 

Find \lim _{ x\rightarrow -2 }{ \cfrac { { x }^{ 2 }+x-2 }{ x+2 }  }  and hence sketch the curve y=\cfrac { { x }^{ 2 }+x-2 }{ x+2 } .

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Substituting x=-2 into \cfrac { { x }^{ 2 }+x-2 }{ x+2 } gives \cfrac{0}{0}.

\lim _{ x\rightarrow -2 }{ \cfrac { { x }^{ 2 }+x-2 }{ x+2 } } \\ \\ =\lim _{ x\rightarrow -2 }{ \cfrac { (x-1)(x+2) }{ x+2 } } \\ \\ =\lim _{ x\rightarrow -2 }{ (x-1) } \\ \\ =-3

y=\cfrac { { x }^{ 2 }+x-2 }{ x+2 } \\ \\ y=\cfrac { (x+2)(x-1) }{ x+2 } \\ \\ y=x-1

So the function y=x-1 where x\neq -2. It is discontinuous at (-2, -3).