Chapter 8: Continuation of Limits


We briefly recall the notion of a limit.

limxCf(x)=L\lim_{x \rightarrow C} f(x) = L \Leftrightarrow  f(x)f(x) is close to LL when xx is close to CC


Calculate limx2(x2+5x2)\lim_{x \rightarrow 2} (-x^2+5x-2)

Because the limit is a simple polynomial we may evaluate by substitution.

limx2(x2+5x2)=(2)2+522\lim_{x \rightarrow 2} (-x^2+5x-2) = -(2)^2+5 \cdot 2 -2 

=4+102=4=-4 +10 -2 = 4


Calculate limz4z210\lim_{z \rightarrow 4} \sqrt{z^2-10}

=4210=1610=6= \sqrt{4^2-10} = \sqrt{16-10}=\sqrt{6}

We need to be careful with direct evaluation.


Calculate limx3x+3x2+4x+3\lim_{x \rightarrow -3} \frac{x+3}{x^2+4x+3}

(3)+3=0(-3)+3=0 and (3)2+4(3)+3=0(-3)^2+4(-3)+3=0

Thus we get "00\frac{0}{0}" which is not a number

limx3x+3x2+4x+3\lim_{x \rightarrow -3} \frac{x+3}{x^2+4x+3} =limx3x+3(x+4)(x+3)=\lim_{x \rightarrow -3} \frac{x+3}{(x+4)(x+3)}

=limx31x+4= \lim_{x \rightarrow -3} \frac{1}{x+4}



One Sided Limits


What is the behaviour of f(x)=xf(x)=\sqrt{x} "close to" x=0x=0?

Consider the graph:

  1. Sloping up when x0x \geq 0
  2. Not defined for x<0x<0


What is the behaviours of f(x)=xxf(x) = \frac{|x|}{x} "close tp" x=0x=0?

Consider the graph:

  1. f(x)=1f(x)=1 when x0x \geq 0
  2. f(x)=1f(x)=-1  when x<0x<0

We observe that there can be very distinct behaviour "on this left" and "on the right".


The right hand limit as xx approaches x=cx=c of f(x)f(x) is

limxC+f(x)=L\lim_{x \rightarrow C^+} f(x) = L

The left hand limit is

limxCf(x)=L\lim_{x \rightarrow C^-} f(x) = L

We obtain limx0+xx=1\lim_{x \rightarrow 0^+} \frac{|x|}{x}=1 and limx0xx=1\lim_{x \rightarrow 0^-} \frac{|x|}{x}=-1


Consider the following graph:

Which of the following exist (are defined)?

limx1+f(x)\lim_{x \rightarrow -1^+} f(x) limx1f(x)\lim_{x \rightarrow -1^-} f(x)

limx3+f(x)\lim_{x \rightarrow 3^+} f(x) limx3f(x)\lim_{x \rightarrow 3^-} f(x)

limx1f(x)\lim_{x \rightarrow 1^-} f(x) limx1+f(x)\lim_{x \rightarrow 1^+} f(x)


The limit limxCf(x)\lim_{x \rightarrow C} f(x) EXISTS if limxcf(x)=limxc+f(x)\lim_{x \rightarrow c^-} f(x) = \lim_{x \rightarrow c^+} f(x)


Pick a value AA so that :

limxf(x)\lim_{x \rightarrow} f(x)  exists when f(x)={3x+2x2 x2+Ax<2f(x)=\left\{\begin{matrix} 3x+2 & x \geq 2\\ \ x^2+A& x<2 \end{matrix}\right.

We need limx2f(x)=limx2+f(x)\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^+} f(x)

limx2f(x)=limx2x2+A\lim_{x \rightarrow 2^-} f(x) = \lim_{x \rightarrow 2^-} x^2 +A \Rightarrow  x2x \rightarrow 2^- gives


limx2+=limx2+3x+2 \lim_{x \rightarrow 2^+}=\lim_{x \rightarrow 2^+} 3x+2 \Rightarrow x2+x \rightarrow 2^+ gives x2x \geq 2


Thus, 8=4+A8=4+A and A=4A=4

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