Limits Formula

Calculus is a very interesting and application-oriented branch of mathematics. It deals with many mathematical principles and theorems which are very helpful in deriving various conclusions and results in science and technology. Limit theory is the most fundamental and important concept of calculus. It deals with the determination of values at some point, which may not be deterministic exactly otherwise. In this article, we will discuss some important Limits Formula and their examples. Let us learn the concept.

Limits Formula

What is Limit?

Let y = f(x) as a function of x.  If at a point x = a, f(x) takes indeterminate form, then we can consider the values of the function which is very near to a. If these values tend to some definite unique number as x tends to a, then that obtained a unique number is called the limit of f(x) at x = a. We can write it.

limx→af(x)

For example

limx→2f(x)=5

Here, as x approaches 2, the limit of the function f(x) will be 5i.e. f(x) approaches 5. The value of the function which is limited and can be different than the limit value of the function itself. Also, we can see that a function value may or may not be the same as the limit and that either value may be undefined.

The limit value is having two types of values as Left-hand Limit and Right-Hand Limit.

If values of the function at the points, very close to a on the left tends to a definite unique number as x tends to a. Then the unique number, such obtained is called the left hand limit of f(x) at x = a. We write it as

limx→af(x)

Left Hand Limit :

f(a-0)= limx→a−f(x)=limh→0f(a−h)

Similarly, Right Hand Limit:

f(a+0)= limx→a+f(x)=limh→0f(a+h)

Now, Limit will exist at x=a if following two conditions are satisfied.

(i) limx→a−f(x)andlimx→a+f(x) both exist.

(ii) limx→a−f(x)=limx→a+f(x)

Uniqueness of Limit: Limit value will be unique, if exists. There cannot be two distinct numbers L1andL2 ,  such that when x tends to a, the given function f(x) tends to both L1andL2.

Some Important Limit Formula

(i) limx→0 sin x = 0

(ii) limx→0cos x = 1

(iii) limx→0sinxx= 1

(iv) limx→0log(1+x)x = 1

(v)limx→0logex = 1

(vi) limx→elogex = 1

(vii) limx→0ex−1x = 1

(viii) limx→0ax−1x=logea

Solved Examples

Q.1: Find value of

limx→0x2+1

Solution: We have,

limx→0x2+1

Put x= 0 directly, we get value of limit as 1.

Q.2: Find the value of,

limx→5x2−5x230

Solution: Since direct substitution of x=5 in the expression, we will get

200

Which is not deterministic. So, we will do the following steps.

First do factorization of both denominator and numerator. We will get-

i.e. limx→5(x+5)(x−5)(x+6)(x−5)

i.e. limx→5(x+5)(x+6)

Now put x= 5 , we have value as,

1011 , which is the answer.