# What are the limits in calculus and how to calculate it algebraically?

In calculus, limits are used to evaluate the numerical value of the function or any type of output instead of infinity by applying different techniques to it. The limit can be evaluated for algebraic, trigonometric, exponential, logarithmic, etc. functions.

There are different techniques for calculating the problems of limits such as substitution, factorization, rationalization, or applying the rules of limits. In this article, we are going to explain the term limits in calculus and how to evaluate the limit of algebraic expressions.

## What are the limits in calculus?

Limit is the sub-branch of calculus that allows a function to reach some specific value as the given function’s expression goes closer and closer to the particular value. The specific value must be given for the independent variable of the function.

The limit must be evaluated with the help of the particular value of the independent variable as the value is substituted to the function. If the function gives zero by zero or infinity form then we have to rationalize, factorize, or use L’hopital’s rule to evaluate the limit of that function.

### Formula of limit

The limit of a function f(z) as z approaches** “a**” is equal to M and is said to be the formula of the limit calculus. Mathematically it can be written as:

**Lim _{z}**

_{→a}**f(z) = M**

**z**= the independent variable of the function**a**= the particular point of the function.**f(z)**= the given function**M**= the numerical result of the function

### Rules of limit in calculus

The basic rules of limit calculus are given below.

Name of the rule | Rule |

Sum Rule | Lim_{z}_{→a} [f(z)+ g(z)] = Lim_{z}_{→a} [f(z)]+ Lim_{z}_{→a} [g(z)] |

Difference Rule | Lim_{z}_{→a} [f(z)- g(z)] = Lim_{z}_{→a} [f(z)]- Lim_{z}_{→a} [g(z)] |

Constant Rule | Lim_{z}_{→a} [k] = k |

Constant function Rule | Lim_{z}_{→a} [k * f(z)] = k * Lim_{z}_{→a} [f(z)] |

Power Rule | Lim_{z}_{→a} [f(z)]^{2} = [Lim_{z}_{→a} [f(z)]^{2}^{} |

Product Rule | Lim_{z}_{→a} [f(z)* g(z)] = Lim_{z}_{→a} [f(z)]* Lim_{z}_{→a} [g(z)] |

Quotient Rule | Lim_{z}_{→a} [f(z)/ g(z)] = Lim_{z}_{→a} [f(z)]/ Lim_{z}_{→a} [g(z)] |

L’hopital’s Rule | Lim_{z}_{→a} [f(z)/ g(z)] = Lim_{z}_{→a} [f’(z)/ g’(z)] |

## How to evaluate the limits algebraically?

To evaluate the limit algebraically, you have to understand how to use rationalization, factorization, or L’hopital’s rule of the function that makes infinity or zero by zero form. Follow the below steps to evaluate the limit algebraically.

- First of all, evaluate the limit of the function.
- If the result of the function makes infinity, infinity by infinity, or zero by zero form then apply the factorization to the given function.
- If the function is unable to factorize then rationalize it.
- If the function is not able to rationalize then use L’hopital’s rule of the limit.

A limits calculator can be used to solve the limit of algebraic expressions.

Let us take a few examples of limits to solve simple and fractional algebraic expressions manually.

### For simple algebraic expression

**Example**

Evaluate the limit of f(z) = 2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5 if the particular value is 2.

**Solution**

**Step 1:** First of all, write the notation of the limit with the given function.

f(z) = 2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5

lim_{z}_{→}_{a} [f(z)] = lim_{z}_{→}_{2} [2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5]

**Step 2:** Now apply the sum and difference rule of limits and apply the limits notation separately to each term of the function.

lim_{z}_{→}_{2} [2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5] = lim_{z}_{→}_{2} [2z^{2}] * lim_{z}_{→}_{2} [5z^{3}] * lim_{z}_{→}_{2} [2z^{5}] / lim_{z}_{→}_{2} [5z^{4}] + lim_{z}_{→}_{2} [5]

**Step 3:** Take out the constant coefficient outside the limits notation.

lim_{z}_{→}_{2} [2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5] = 2lim_{z}_{→}_{2} [z^{2}] * 5lim_{z}_{→}_{2} [z^{3}] * 2lim_{z}_{→}_{2} [z^{5}] / 5lim_{z}_{→}_{2} [z^{4}] + lim_{z}_{→}_{2} [5]

**Step 4:** Now place the articular value z = 2 to the above expression to evaluate the limit.

lim_{z}_{→}_{2} [2z^{2} * 5z^{3} * 2z^{5} / 5z^{4} + 5] = 2 [2^{2}] * 5 [2^{3}] * 2 [2^{5}] / 5 [2^{4}] + [5]

= 2 [2 x 2] * 5 [2 x 2 x 2] * 2 [2 x 2 x 2 x 2 x 2] / 5 [2 x 2 x 2 x 2] + [5]

= 2 [4] * 5 [8] * 2 [32] / 5 [16] + [5]

= 8 * 40 * 64 / 80 + 5

= 320 * 64 / 80 + 5

= 2048 / 80 + 5

= 25.6 + 5

= 30.6

### For fractional algebraic expression

Example 1: for rationalization

Evaluate the limit of f(w) = (w^{2} – 10w + 9)/(w^{2} – 81) if the particular value of the function approaches infinity.

**Solution**

**Step 1:** First of all, write the notation of the limit with the given function.

f(w) = (w^{2} – 10w + 9)/(w^{2} – 81)

lim_{z}_{→}_{a} [f(z)] = lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)]

**Step 2:** Apply the limit to the given expression.

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [(lim_{z}_{→∞} w^{2} – lim_{z}_{→∞} 10w + lim_{z}_{→∞} 9)/( lim_{z}_{→∞} w^{2} – lim_{z}_{→∞} 81)]_{}

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [(∞^{2} – 10(∞) + 9)/( ∞^{2} – 81)]_{}

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [(∞)/( ∞)]

it forms infinity by infinity.

**Step 3:** Let us rationalize the given function.

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [(w^{2} – 9w – w + 9)/(w^{2} – 9^{2})]_{}

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [(w(w – 9) –1 (w – 9))/(w^{2} – 9^{2})]_{}

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [((w – 9) (w – 1))/((w – 9) (w + 9))]

**Step 4:** Now cancel the similar terms in the numerator and denominator and take “w” as a common from the remaining terms_{}

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [(~~(w – 9) ~~(w – 1))/(~~(w – 9) ~~(w + 9))]

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [(w – 1) / (w + 9)]

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [~~w~~(1 – 1/w) / ~~w~~(1 + 9/w)]

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = lim_{z}_{→∞} [(1 – 1/w) / (1 + 9/w)]

Step 5: Now apply the limit.

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [(1 – 1/∞) / (1 + 9/∞)]

As something over infinity always gives zero so,

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [(1 – 0) / (1 + 0)]

lim_{z}_{→∞} [(w^{2} – 10w + 9)/(w^{2} – 81)] = [1/1] = 1

**Example 2: For L’hopital’s rule**

Evaluate the limit of f(t) = t^{3} – 2t^{2} + 1 / (4t^{2} – 4) if the particular value is 1.

**Solution**

**Step 1:** First of all, write the notation of the limit with the given function.

f(t) = t^{3} – 2t^{2} + 1 / (4t^{2} – 4)

lim_{t}_{→}_{a} [f(t)] = lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)]

**Step 2:** Apply the limit to the given expression

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [(lim_{t}_{→1} t^{3} – lim_{t}_{→1} 2t^{2} + lim_{t}_{→1} 1) / (lim_{t}_{→1} 4t^{2} – lim_{t}_{→1} 4)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [(1^{3} – 2(1)^{2} + 1) / (4(1)^{2} – 4)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [(1 – 2 + 1) / (4(1) – 4)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [(– 1 + 1) / (4 – 4)] = 0/0

**Step 3:** Apply L’hopital’s rule of limit as the function gives 0/0 form. Now find the derivative of the function and apply the limit value again.

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = lim_{t}_{→1} [d/dt [t^{3} – 2t^{2} + 1] / d/dt (4t^{2} – 4)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = lim_{t}_{→1} [[3t^{2} – 4t + 0] / (8t – 0)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = lim_{t}_{→1} [[3t^{2} – 4t] / (8t)]

Now substitute t = 1

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [[3(1)^{2} – 4(1)] / (8(1))]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [[3(1) – 4(1)] / (8(1))]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = [[3 – 4] / (8)]

lim_{t}_{→1} [t^{3} – 2t^{2} + 1 / (4t^{2} – 4)] = -1/8 = -0.125

## Conclusion

Now you can solve any problem of the limits in calculus by following the above post. The limit of algebraic expression can be evaluated easily as mentioned above. You can grab all the basics of this post by learning and practicing it.