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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 $$\delta$$

#### 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

A

0.5

.

B

–0.8

C

0.1

D

7

Option C is Correct

#### 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 $$|x^2-4|<\dfrac {1}{3}$$

A

0.082

.

B

4

C

–1

D

3

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$$f(x) < N$$

#### 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 $$|f(x)-\ell|<\epsilon$$

#### 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

# 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 $$|f(x)-\ell|<\epsilon$$

#### 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 $$f(x) > M$$.

#### 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