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Precise Definition Of Limits

Learn precise definition of a limit & one sided infinite limits, practice calculus one sided limit problems & solutions.

Precise Definition of Limit

The Meaning of 

\(\lim\limits_{x\rightarrow a}\;f(x)\) was the values of 'f' as x takes values closer and closer to 'a'. Now how close?

So this definition seems a little vague.

  • We write \(\lim\limits_{x\rightarrow a}\;f(x)=\ell\) if for every number \(\epsilon>0\) there is a number \(\delta >0\) such that if \(0<|x-a|<\delta\), then \(|f(x)-l|<\epsilon\).
  • Another meaning of this is that the values of f(x) can be made as close to \('\ell'\) as we desire by taking x close to 'a'.

 

\(\lim\limits_{x\rightarrow a}\;f(x)=\ell\)

where, x is in the range.

\(a-\delta<x<a+\delta\)

then f(x) is in the range \(\ell-\epsilon<f(x)<\ell+\epsilon\)

  • Equivalently if x lies in \((a-\delta,\; a+\delta)\), then f(x) lies in \((\ell-\epsilon,\; \ell+\epsilon)\)
  • From \(|f(x)-l|<\epsilon\) ,we find the range in which x lies, because we know f(x), \(\ell\)and \(\epsilon\).
  • Now compare this range with that obtained from \(|x-a|<\delta\) and obtain the range for \(\delta\)

Illustration Questions

Using the given graph of 'f' find a number \(\delta\)such that \(|x-2|<\delta\), then | f(x) –3| < 0.3

A 0.5

B –0.8

C 0.1

D 7

×

\(|f(x)-3|<0.3\)

\(\Rightarrow -0.3< f(x)-3<0.3\)

\(\Rightarrow2.7<f(x)<3.3\)  (Look at the graph when the height is between 2.7 and 3.3 the corresponding values of x are between 1.9 and 2.2)

So, \(\Rightarrow2.7<f(x)<3.3\)

\(\Rightarrow1.9<x<2.2\)

or

Now this interval (1.9, 2.2) is not symmetric about x = 2. The distance of left endpoint 1.9 is

2 –1.9 = 0.1 and that of right end points is 2.2 – 2 = 0.2.

Choose \(\delta\)to be smaller of these two .

\(\therefore\) \(\delta=0.1\) (or any smaller number than this)

Using the given graph of 'f' find a number \(\delta\)such that \(|x-2|<\delta\), then | f(x) –3| < 0.3

image
A

0.5

.

B

–0.8

C

0.1

D

7

Option C is Correct

Illustration Questions

Use the given graph of \(f(x)=x^2\),  find value of \(\delta\)such that if \(|x-2|<\delta\), then \(|x^2-4|<\dfrac {1}{3}\)

A 0.082

B 4

C –1

D 3

×

\(|x^2-4|<\dfrac {1}{3}\)

\(\Rightarrow \dfrac {-1}{3}<x^2-4<\dfrac {1}{3}\)

\(\Rightarrow \dfrac {11}{3}<x^2<\dfrac {13}{3}\)

Now the values of  x' corresponding to \(x=\dfrac {11}{3}\) and \(x=\dfrac {13}{3}\) are

\(x^2=\dfrac {11}{3} \Rightarrow x=\sqrt {\dfrac {11}{3}}\) = 1.915 and 

\(x^2=\dfrac {13}{3} \Rightarrow x=\sqrt {\dfrac {13}{3}}\)= 2.082

Therefore, \(x\in\) (1.915, 2.082), when \(x^2\in \left ( \dfrac {11}{3},\dfrac {13}{3} \right)\).

The interval is not symmetric about x = 2, choose \(\delta\)to be

min {2 –1.915, 2.082 – 2} = min {0.085, 0.082}

\(\Rightarrow \delta=0.082\) (any values less than this will also work)

Use the given graph of \(f(x)=x^2\),  find value of \(\delta\)such that if \(|x-2|<\delta\), then \(|x^2-4|<\dfrac {1}{3}\)

image
A

0.082

.

B

4

C

–1

D

3

Option A is Correct

Left Hand Limits (Precise definition)

  • We say that \(\lim\limits_{x \rightarrow a^-}\;f(x)=\ell\). If for every number \(\epsilon>0\) there is a number \(\delta>0\) such that if \(a-\delta<x<a\), then \(|f(x)-\ell|<\epsilon\)

Illustration Questions

If \(\lim\limits_{x\rightarrow 2^-}\;(x^2-x-3)=-1\), find the value of \(\delta\) that corresponds to \(\epsilon=0.1\) in the definition.

A 0.025

B 10

C 8

D 17

×

 \(\lim\limits_{x \rightarrow a^-}\;f(x)=\ell\) 

⇒ For every number \(\epsilon>0\) there is a number \(\delta>0\) such that if \(x\in(a-\delta,a)\), then \(|f(x)-\ell|<\epsilon\)

In this problem,

\(\lim\limits_{x \rightarrow 2^-}\;(x^2-x-3)=-1\)

\(\therefore\) |x2 –x – 3 –(–1)| < 0.1

\(\Rightarrow |x^2-x-2|<0.1\)

\(\Rightarrow|(x-2)(x+1)|<0.1\)

\(\Rightarrow |x-2|<\dfrac {0.1}{|x+1|}\)

Now, \(2-\delta < x < 2\)  ...(i)

 and 

\( |x-2|<\dfrac {0.1}{|x+1|}\)

\(\Rightarrow |x-2|<\dfrac {0.1}{4}\)               (Take \(|x+1|<4\) if x is around 2. Take a margin of 1)

\(\Rightarrow |x-2|<0.025\)

Comparing with (i) we get, \(\delta = 0.025\)

Choose \(\delta = 0.025\) or any value less than this.

If \(\lim\limits_{x\rightarrow 2^-}\;(x^2-x-3)=-1\), find the value of \(\delta\) that corresponds to \(\epsilon=0.1\) in the definition.

A

0.025

.

B

10

C

8

D

17

Option A is Correct

Positive Infinite Limits \((+\infty)\)

We say,

\(\lim\limits_{x\rightarrow a}\;f(x)=\infty\) , if for every positive number M there is a positive number \(\delta\)such that if \(0<|x-a|<\delta\), then \( f(x) > M\).

Illustration Questions

If \(\lim\limits_{x\rightarrow -2}\;\dfrac {1}{(x+2)^2}=\infty\), find the value of \(\delta\) for M = 10000 in the definition.

A 0.01

B 5

C 1.3

D –1.7

×

\(\dfrac {1}{(x+2)^2}>10000\)

\(\Rightarrow (x+2)^2<\dfrac {1}{10000}\)

\(\Rightarrow (x+2)^2<0.0001\)

\(\Rightarrow-0.01<x+2<0.01\)

\(\Rightarrow-2.01<x<–1.98\)

\(\Rightarrow-0.01<x+2<0.01\)

\(\Rightarrow |x+2|<0.01\)

So choose \(\delta=0.01\)   or can take any value less than this.

If \(\lim\limits_{x\rightarrow -2}\;\dfrac {1}{(x+2)^2}=\infty\), find the value of \(\delta\) for M = 10000 in the definition.

A

0.01

.

B

5

C

1.3

D

–1.7

Option A is Correct

Negative Infinite Limits \((-\infty)\)

We say

\(\lim\limits_{x\rightarrow a}\;f(x)=-\infty\), if for every negative number N there is a positive number \(\delta\) such that if \(0<|x-a|<\delta\), then\( f(x) < N\)

Illustration Questions

If \(\lim\limits_{x\rightarrow -2^-}\;\dfrac {1}{(x+2)^3}=-\infty\), then find the value of \(\delta\) for N = –1000000 in the definition.

A 0.01

B 5

C 17

D –9

×

\(\dfrac {1}{(x+2)^3}<-10^{6}\)

\(\Rightarrow (x+2)^3>-10^{-6}\) (Take reciprocals)

\(\Rightarrow x+2>-10^{-2}\) (Take cube root on both sides)

\(\Rightarrow|x+2|<10^{-2}\)

\(\Rightarrow|x+2|<0.01\)

\(\therefore \;\delta=0.01\)or can take value less than this.

 

If \(\lim\limits_{x\rightarrow -2^-}\;\dfrac {1}{(x+2)^3}=-\infty\), then find the value of \(\delta\) for N = –1000000 in the definition.

A

0.01

.

B

5

C

17

D

–9

Option A is Correct

Right Hand Limits(Precise definition)

  •  We say that \(\lim\limits_{x \rightarrow a^+}\;f(x)=\ell\) If for every number \(\epsilon>0\) there is a number \(\delta>0\) such that if \(a<x<a+\delta\), then \(|f(x)-\ell|<\epsilon\)      

Illustration Questions

If \(\lim\limits_{x\rightarrow 2^+}\;(x^2-x-3)=-1\), find the value of \(\delta\)that corresponds to \(\epsilon=0.1\) in the definition.

A 0.02

B 10

C 8

D 17

×

 \(\lim\limits_{x \rightarrow a^+}\;f(x)=\ell\)  ,for every number \(\epsilon>0\) there is a number \(\delta>0\) such that if \(x\in(a,\; a+\delta)\), then \(|f(x)-\ell|<\epsilon\)

In this problem,

\(\lim\limits_{x \rightarrow 2^+}\;(x^2-x-3)=-1\)

\(|x^2 –x – 3 –(–1)| < 0.1\)

\(\Rightarrow |x^2-x-2|<.1\)

\(\Rightarrow|(x-2)(x+1)|<.1\)

\(\Rightarrow |x-2|<\dfrac {0.1}{|x+1|}\)

Now, \( 2 < x < 2 +\delta\)  ...(i)

 and 

\(\Rightarrow |x-2|<\dfrac {0.1}{|x+1|}\)

\(\Rightarrow |x-2|<\dfrac {0.1}{4}\)               (Take \(|x+1|<4\) if x is around 2. Take a margin of 1)

\(\Rightarrow |x-2|<0.025\)

Comparing with (i) we get, \(\delta = 0.025\)

Choose \(\delta = 0.025\) or any value less than this.

If \(\lim\limits_{x\rightarrow 2^+}\;(x^2-x-3)=-1\), find the value of \(\delta\)that corresponds to \(\epsilon=0.1\) in the definition.

A

0.02

.

B

10

C

8

D

17

Option A is Correct

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