Limit in math : All Formula And sum explanation [tips and trick]

How To Use Limit in Mathematics

• Limit Table Of content
• Limit Definition In Maths
• Properties of Limit Formula
• Type1:How To Put Direct Limit
• Type2: 0/0 Form (factorisation) solution
• Type3:  Sum of Root (√) type
• Type4:  Sum of 1/n form or infinity form
• Limit in maths Type 5 Formula
• Limit in maths Type 6 Formula
• Limit in maths Type 7 Formula
• Limit in maths Type 8 Formula

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Limit Definition In Maths

What is Limit Definition

Let us consider a real-valued function “f” and the real number “c”, the limit is normally defined as

`\lim_{x \rightarrow c}`f(x) = L

It is read as “the limit of f of x, as x approaches c equals L”. The “lim” shows limit, and fact that function f(x) approaches the limit L as x approaches c is described by the right arrow.

What is Limit Definition

Let us consider a real-valued function “f” and the real number “c”, the limit is normally defined as

`\lim_{x \rightarrow c}`f(x) = L

It is read as “the limit of f of x, as x approaches c equals L”. The “lim” shows limit, and fact that function f(x) approaches the limit L as x approaches c is described by the right arrow.

All Formula for limit
( Properties of Limit)

01) `\lim_{x \rightarrow a}` (k) = k

02) `\lim_{x \rightarrow a}` k f(x) = k `\lim_{x \rightarrow a}` f(x)

03) `\lim_{x \rightarrow a}` [f(x) + g(x)] = `\lim_{x \rightarrow a}` f(x) + `\lim_{x \rightarrow a}` g(x)

04) `\lim_{x \rightarrow a}` [f(x) - g(x)] = `\lim_{x \rightarrow a}` f(x) - `\lim_{x \rightarrow a}` g(x)

05) `\lim_{x \rightarrow a}` [f(x) • g(x)] = `\lim_{x \rightarrow a}` f(x) • `\lim_{x \rightarrow a}` g(x)

06) `\lim_{x \rightarrow a}` `\frac{f(x)}{g(x)}` = `\frac{\lim_{x \rightarrow a} f(x)}{\lim_{x \rightarrow a} g(x)}`

07) `\lim_{x \rightarrow a}` log f(x) = log[`\lim_{x \rightarrow a}` f(x)]

Type1:How To Put Direct Limit

Put The Direct Limit

Note

When you put Direct limit? , Which Type of sums put direct limit?

Put Limit in your question(rought page), Check if your answer 0/0, x/0 (denometer 0), When this type of case then you did not put direct limit,in 0/x,0/0 this type of cases, simplify sum and then put limit(This type of case we will discuss in type2).

Ex:1  `\lim_{x \rightarrow 1}` `\frac{x^{2} + 4x +2}{x +1}`

Google search:  limit of x tends to 1 `x^{2}` +4x +2 / x+1

How to calculate direct limit sum

step 1: 
Put limit in your question(raf page), If your is not 0/0 form, then put direct limit (step 1 reason will be note session.)
= `\frac{1^{2} + 4(1) + 2}{1 + 1}`

step 2: 
Answer = `\frac{7}{2}`

Type2: 0/0 Form (factorisation) solution

How to solution 0/0 Form (factorisation)

Note

Limits by factorization method is applicable when the value of limit gives you any of these 0/0 form then we will use the factorization method to evaluate that limit.

It is an expression involving two functions whose limits can not be determined.

the limit obtained by direct substitution is of the form 0/0.

Ex:1  `\lim_{x \rightarrow 1}` `\frac{x^{2} -4x +3}{x^{2} +2x -3}`

Google search:  limit of x tends to 1 `x^{2}` -4x +3 / `x^{2}` +2x -3

step 1: 
limit of x tends to 1 `x^{2}` -4x +3 / `x^{2}` +2x -3 if x= 1 in the given function then we will get which is an 0/0 form. So we will solve this limit by factorization method. = `\frac{(x -3) (x -1)}{(x +3) (x-1)}`

step 2: 
Put x limit=  `\frac{1 -3}{1 +3}`

step 3: 
Answer =  `\frac{-1}{2}`

0/0 Form (factorisation) in extra sum

1)  `\lim_{x \rightarrow 2}` `\frac{x^{4} -8x^{2} +16}{x^{3} -3x^{2} +4}`

1)Google search:  limit of x tends to 2 `x^{4}` -`8x^{2}` +16 / `x^{2}` -`3x^{2}` +4

Answer:- `\frac{16}{3}`

2)  `\lim_{x \rightarrow 2}` `\frac{x^{3} -x^{2} -5x +6}{x^{2} -5x +6}`

2)Google search:  limit of x tends to 2 `x^{3}` -`x^{2}` -5x +6 / `x^{2}` -5x +6

Answer:- 3

3)  `\lim_{x \rightarrow -3}` `\frac{x^{3} +27}{x^{2} +5x +6}`

3)Google search:  limit of x tends to -3 `x^{3}` +27 / `x^{2}` +5x +6

Answer:- 3

Type3: What is sum of root (√) type

sum of root (√) type in limit

Ex:1  `\lim_{x \rightarrow 0}` `\frac{\sqrt{9+x} -3}{x}`

Google search:  limit of x tends to 0 `\sqrt{9+x}` -3 / x

step 1: 

In This we can see that this limit is tending towards 0/0 form so we are going to rationalize in terms of numerator and we will get it as,`\Rightarrow` `\lim_{x \rightarrow 0}` = `\frac{\sqrt{9+x} - 3}{x}\times\frac{\sqrt{9+x} + 3}{\sqrt{9+x} + 3}`

step 2: 

Now when we see the numerator and we can see that the form that is formed is of (x - y)(x + y) = `x^2` - `y^2`so we can also write the numerator as

`\Rightarrow` `\lim_{x \rightarrow 0}` = `\frac{9 + x - 9}{x (\sqrt{9+x} + 3)}`

`\Rightarrow` `\lim_{x \rightarrow 0}` = `\frac{1x}{x (\sqrt{9+x} + 3)}`

= `\frac{1}{ (\sqrt{9+0} + 3)}`

= `\frac{1}{3 + 3}`

Answer =  `\frac{1}{6}`

Sum of root (√) type in limit extra sum

1)  `\lim_{x \rightarrow 0}` `\frac{\sqrt{1-x} - \sqrt{1+x}}{x}`

1)Google search:  limit of x tends to 0 √1-x - √1+x / x

Answer:- `-1`

2)  `\lim_{n \rightarrow \infty}` `(\sqrt{n^2 +n +1} -n)`

2)Google search:  limit of n tends to infity `(\sqrt{n^2 +n +1} -n)`

Answer:-  1/2

Type4:  Sum of 1/n form or infinity form

What is Sum of 1/n form or Limit of infinity Formula

01) `\lim_{n \rightarrow \infty}` • `\frac{1}{n}` = 0

02) `\lim_{n \rightarrow \infty}` `r^{n}` = 0

03) `\lim_{0 \rightarrow 0}` `\frac{sinθ}{θ}` = 1 `\Rightarrow`  `\lim_{0 \rightarrow 0}` `\frac{sin^{-1}x}{x}` = 1

04) `\lim_{0 \rightarrow 0}` `\frac{tanθ}{θ}` = 1 `\Rightarrow`  `\lim_{0 \rightarrow 0}` `\frac{tan^{-1}x}{x}` = 1

05)  ∑ 1   = n `\Rightarrow` ∑   `n^{2}` = `\frac{n(n +1)(2n +1)}{6}`

06)  ∑ n   = `\frac{n(n+1)}{2}` `\Rightarrow` ∑   `n^{3}` = `\frac{n^{2}(n+1)^{2}}{4}`

note

After read all formulas learn this sums.

Ex:1  `\lim_{n \rightarrow \infty}` `\frac{4n -6}{2n +5}`

Google search:  limit of x tends to infinity 4n -6 / 2n +5

step 1: 

See question, try to set 1/n formula in question. In this case we devide n/n in question like that.
`\Rightarrow` `\frac{\frac{4n-6}{n}}{\frac{2n +5}{n}}`

step 2: 

Now when we see the numerator and we can see that the form that is formed is of (a - b/n) / (a + b/n)so we can also write the numerator as
`\Rightarrow` `\lim_{x \rightarrow \infty}` = `\left(\frac{4-\frac6n}{2+\frac5n}\right)`

= `\frac{4-0}{2+0}`

= `\frac{4}{2}`

Answer =  2

extra sum Sum of 1/n form or infinity Formula

1)  `\lim_{x \rightarrow \infty}` `\frac{1^{2} +2^{2} +3^{2} +..........+n^{2}}{x^3}`

1)Google search:  limit of x tends to infity `1^{2} +2^{2} +3^{2} +..........+n^2` / `x^3`

Answer:- 1/3

2)  `\lim_{n \rightarrow \infty}` `\frac{\sqrt{n^{2} +n +1} -n}{\sqrt{n^{2} +n +1} -n}`

2)Google search:  limit of n tends to infity `\sqrt{n^2 +n +1}` -n / `\sqrt{n^2 +n +1}` -n

Answer:-  1/2

3)  `\lim_{x \rightarrow \infty}` `\left(\frac{n+1}{n+2}\right)^n`

3)Google search:  limit of x tends to infity n+1/n+2 / n

Answer:- 1/e

Limit in maths Type 5 Formula

What is type 5 Formula in limit

 `\lim_{x \rightarrow a}`  `\frac{x^n -a^n}{x -a}` = `na^{n-1}`

Ex:1  `\lim_{x \rightarrow 2}` `\frac{x^3 -8}{x -2}`

Google search:  limit of x tends to 2 `x^3` -8 / x -2

step 1: 

see your numereter first term in this case `x^3` is numereter, numereter second term is realeted to first term, first term is `x^3` term set your second term is qube formate. In this case 8 is set with `2^3` 8 = `2^3`
`\Rightarrow` `\frac{x^3 -2^3}{x -2}`

step 2: 

Put the formula.
`\Rightarrow` `3(2)^{3-1}`

= `3(2)^2`

= 3 × 4

Answer =  12

Type 5 extra sum in limit

1)  `\lim_{x \rightarrow 2}` `\frac{x^3 -8}{x^2 -2}`

1)Google search:  limit of x tends to 2 `x^3` -8 / `x^2` -2

Answer:- 3

2)  `\lim_{n \rightarrow 3}` `\frac{x^3 -27}{`\sqrt[3]{x}` - `\sqrt[3]{3}`}`

2)Google search:  limit of n tends to 3 `x^3` -27 / `\sqrt[3]{x}` - `\sqrt[3]{3}`

Answer:-  81`\sqrt[3]{9}`

Limit in maths Type 6 Formula

What is type 6 Formula in limit

 `\lim_{x \rightarrow 0}`  `\frac{a^x -1}{x}` = `\log_{e}a`

Ex:1  `\lim_{x \rightarrow 0}` `\frac{4^x -3^x}{x}`

Google search:  limit of x tends to 0 `4^x -3^x` / x

step 1: 

See the numereter, we have require -1 , +1 to set a formula. In this case we add -1 , +1 then no change in your question but set your formula.
`\Rightarrow` `\lim_{x \rightarrow 0}` `\frac{4^x -1 -3^x -1}{x}`

step 2: 

Accroding to your formula Devide separate term `4^x` -1/x - `3^x` -1/x
`\Rightarrow` `\lim_{x \rightarrow 0}` `\left[\frac{\left(4^x-1\right)}x\-\frac{\left(3^x-1\right)}x\right]`

step 3: 

Accroding to your formula Devide separate term `\lim_{x \rightarrow 0}` `4^x` -1/x `\lim_{x \rightarrow 0}` `3^x` -1/x
`\Rightarrow` `\lim_{x \rightarrow 0}` `\frac{4^x -1}{x}` `\lim_{x \rightarrow 0}` `\frac{3^x -1}{x}`

=log4 - log3

Answer =  log `\frac{4}{3}`

Type 6 extra sum in limit

1)  `\lim_{x \rightarrow \infty}` x(x√5-1)

1)Google search:  limit of x tends to infinity x(x√5 -1)

Answer:-  `\log_{e}a`

2)  `\lim_{n \rightarrow 0}` `\frac{4^x -3^x}{x}`

2)Google search:  limit of n tends to 0 `4^3` -`3^x` / x

Answer:-  log`\frac{4}{3}`

Limit in maths Type 7 Formula

What is type 7 Formula in limit

 `\lim_{n \rightarrow \infty}`  `(1 + \frac{1}{n})^n` = e

 `\lim_{n \rightarrow 0}`  `(1 + n)^frac{1}{n}` = e

Ex:1  `\lim_{x \rightarrow 0}` `(1 +3x)^frac{1}{x}`

Google search:  limit of x tends to 0 (1 +3x)res to 1/x

`\Rightarrow` `\lim_{x \rightarrow 0}` `[(1 +3x)^frac{1}{3x}]^frac{3x}{x}`
Answer =  `e^3`

Ex:2  `\lim_{x \rightarrow 0}` `(1 +\frac{3x}{4})^frac{5}{x}`

Google search:  limit of x tends to 0 (1 +3x/4)res to 5/x

`\Rightarrow``\lim_{x \rightarrow 0}` `[(1 +\frac{3x}{4})^frac{4}{3x}]^frac{15}{4}`
Answer =  `e^frac{15}{4}`

Type 7 extra sum in limit

1)  `\lim_{x \rightarrow 0}` `(1 -\frac{2x}{3})^frac{1}{x}`
1)Google search:  limit of x tends to 0 (1 - 2x/3)res to 1/x

Answer:-  `e^frac{-2}{3}`

Limit in maths Type 8 Formula

What is type 8 Formula in limit

Ex:1  `\lim_{θ \rightarrow 0}` `\frac{sinθ}{tan3θ}`

Google search:  limit of θ tends to 0 sinθ / tan3θ

step 1: 

See your question, We know that θ/θ is one then set your question θ/θ form. In this case nemeretor and denometer, Dividing the numerator by 4θ and Dividing the denominator by 3θ.

`\Rightarrow`  `\frac{\left[\lim_{\theta\rightarrow0}\frac{\sin4\theta}{4\theta}\right]\times\4\theta}{\left[\lim_{\theta\rightarrow0}\frac{\tan3\theta}{3\theta}\right]\times\3\theta}`

=   `\frac{1×4}{1×3}`

=    `frac{4}{3}`

Type 8 extra sum in limit

1)  `\lim_{x \rightarrow 0}` `frac{2sinx\theta -sin2\theta}{\theta^3}`
1)Google search:  limit of x tends to 0 2sinxθ - sin2θ / θ res to 3

Answer:-  1

2)  `\lim_{θ \rightarrow 0}` `\frac{1 -cos\theta}{\theta^2}`
2)Google search:  limit of θ tends to 0 1 -cosθ / θ res to 2

Answer:-  `frac{1}{2}`

3)  `\lim_{x \rightarrow \frac{π}{4}}` `\frac{2 -sec^2x}{1 -tanx}`
3)Google search:  limit of x tends to π denominator 4 2 - sec x res to 2 / 1 -tanx

Answer:-  2