The Calculation Of Limits

(x²+3x+2)/(x+1) = (x+2)(x+1)/(x+1)= x+2

Limit exists where the function approaches one single value from both the left hand and the right hand side. Only in option B, the function approaches 2 different values as x approaches -3 from LHS and RHS.

At x = 2, the given function becomes undefined as it takes the form 0/0 and hence no particular value will exist at x =2

At x = 3, the value of the function is 1
At x = 4, the value of the function is 1

At x = 1, the value of the function is -1
At x = 0, the value of the function is -1
At x = -1, the value of the function is -1
Hence for all values of x >2 the value ofthe function is 1 and for all values of x <2 the value of function is -1

In option A, at x = 2, the values of y is changing from -1 to 1
In option B, at x =2, the hollow dots shold be present as the function is not defined
In option C, the graph doesn’t show any values for x between 1 to 3

0- is any number that is approaching to zero from the left hand side of zero on the number line but is not zero. So, 0- can be -0.01, -0.001, -0.00001, and so on.

When you divide a number by another number which is very small then the quotient is a very big number. For example, if you divide 6 by 0.000001 then the answer is 6000000. If you keep decreasing the denominator the quotient will become bigger an bigger.

When you divide a number by another number which is very small then the quotient is a very big number. For example, if you divide 6 by 0.000001 then the answer is 6000000. If you keep decreasing the denominator the quotient will become bigger an bigger.

When you divide a number by another number which is very small then the quotient is a very big number. For example, if you divide 5 by 0.000001 then the answer is 5000000. If you keep decreasing the denominator the quotient will become bigger an bigger.