Did you know limits can be used to calculate the time a chemical reaction takes to complete? Even various physics concepts use calculus limits as the base method to solve complex equations. In short, these limits have real-life applications as well. Which means you really need to learn how to solve them. Since we are learning limits, here is a direct method to solve these problems. They aren’t really as hard as they seem to be.

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## Direct Method

Now that we know what limits are let us have a look at the ways to solve these functions. Limits have something called as the direct method in order to solve the different types of function f(x). The different types of functions include polynomial function, rational function, and trigonometric functions.

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To findÂ lim_{xâ†’a}f(x), we first substitute x=a and find the value of f(a). The further procedure depends upon the value of f(a). We will come across the following cases.

**Case 1:Â**If f(a)=0/k (where k is a non zero constant), then we need not solve further. The solution isÂ lim_{xâ†’a}f(x)=0**Case 2:Â**If f(a)=k/0 (where kÂ is a non zeroÂ constant), then we need not solve further. The solution is lim_{xâ†’a}f(x) tends to âˆž**Case 3:Â**If f(a) =k,Â (where kÂ is any constant), then we need not solve further. The solution isÂ lim_{xâ†’a}f(x)=k.**Case 4:Â**If f(a)=0/0 (or any one of the seven indeterminate forms), then we need to solve further. There is no direct method to solve these functions and hence we need to use different methods to solve them.

To solve further in case 4, we can use one from among the following methods:

- Standard Simplification
- Sandwich Theorem
- Trigonometric Simplification
- Exponential Simplification
- Logarithmic Simplification
- L’hopitals Rule
- Simplification of Special functions

But let us keep these for some other day. Now have a look at the solved examples of the direct method for solving the limit functions

## Examples Depending on the Type of Functions

**Limit of a Rational Function By Direct Substitution**

EvaluateÂ lim_{xâ†’1} (xÂ²+1)/(x+100)

Solution: We first put Â x=1 Â in (xÂ²+1)/(x+100)Â ,

âˆ´lim_{xâ†’1Â }(xÂ²+1)/(x+100)

=(1Â²+1 )/(100+1)

=2/101

**Limits of Trigonometric Functions Using Direct Substitution**

EvaluateÂ lim_{xâ†’0} (sinx+2) Â²

Solution: We first putÂ x=0Â Â inÂ (sinx+2) Â²

âˆ´lim_{xâ†’0} (sinx+2)Â²

=0+2 Â²

=4

**Limit of a Polynomial Function**

EvaluateÂ limxâ†’1(xÂ³âˆ’xÂ²+1)

Solution: We first putÂ x=1 Â inÂ xÂ³âˆ’xÂ²+1

âˆ´lim_{xâ†’1} ( xÂ³âˆ’xÂ²+1 ) =1âˆ’1+1

âˆ´lim_{xâ†’1Â }(xÂ³âˆ’xÂ²+1 )=1

## Solved Examples for You

**Question 1: Determine the type of function and solve using the appropriate method:Â lim _{xâ†’2}Â (xÂ²+5x+6 )/(Â 2xÂ²âˆ’3x)**

**Answer :** Both the denominator and the numerator are non-zero algebraic polynomials, so we can directly put x=2

âˆ´lim_{xâ†’2}(xÂ²+5x+6)/(2xÂ²âˆ’3x)

=(2Â²+5Ã—2+6)/(2Ã—2Â²âˆ’3Ã—2)

=(4+10+6)/(8âˆ’6)

=20/2

=10

**Question 2: What is the use or application of the direct method in mathematics?**

**Answer:** In the calculus of the differences, a topic in the mathematics subject, the â€˜direct methodâ€™ is a general technique for creating a proof for the presence of a minimizer for a given functional, presented by â€˜Zarembaâ€™ and â€˜David Hilbertâ€™ in around â€˜1900â€™. The technique relies on the methods of practical analysis and topology.

**Question 3: What are the rules for the limit?**

**Answer:** The rule for the limit states that â€˜the limit of the totality of 2 functions is equivalent to the summation of their limits: â€˜limx â†’ a[f(x)+g(x)] = limx â†’ af(x) + limx â†’ ag(x)â€™.

**Question 4: Writethe formula for the assumed mean method.**

**Answer:** Formula:Here, â€˜aâ€™ = assumed mean.

â€˜fiâ€™ = frequency of ith class.

â€˜diâ€™ = â€˜xi â€“ aâ€™ = deviation of ith class.

**Question 5: What are some laws of limits in mathematics?**

**Answer:**Laws of Limits:

- The Constant Law = â€˜limxâ†’ak=kâ€™.
- The Identity Law = â€˜limxâ†’ax=aâ€™.
- The Addition Law = â€˜limxâ†’af(x)+g(x)=limxâ†’af(x)+limxâ†’ag(x)â€™.
- The Subtraction Law =â€˜limxâ†’af(x)âˆ’g(x)=limxâ†’af(x) â€“ limxâ†’ag(x)â€™, etc.

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