Limits-Examples



                We are going to see some important examples in this page 'Limits-examples'. 

Example1:

Find the limit  x/x+1

Solution:

       The given sequence is 1/2,2/3, 3/4, 4/5, .....999/1000, ...

As the number going on increasing,  that is the value of x is increasing the value of the terms approaches 1. So the value of the limit is 1.

Example 2:

Find the limit as x approaches 0 of (sin x)/x.

Solution:

     We can not directly substitute the value of x as 0 because

sin 0 /0 is identified. We can substitute values of x which is closer to 0 and we can finally conclude that 

             limit  (sin x)/x  = 1.

We will see some types of evaluating the limit problems in the page 'Limits-examples'.

Type I:

Solving limit problems by direct substitution:

1. Evaluate the limit as x tends to a of x².

Solution:  Lim ₓ→ₐ x²  =  a².

(Here we had substitute the value of x as a).


2. Evaluate the limit as x tends to 3 of (x²-10).

Solution: Lim x→₃ (x²-10).

                             =  (3² -10)

                             =   9 - 10

                             =     -1

Type 2:

Solving Limit problems by factoring:

1. Evaluate the limit as x tends to 1 of  (x²-1)/(x-1)

Solution:

                        If we directly apply as x approaches 1, then we will have 0/0.   But if we factorize the   given function, then we will get the limit.

Let us factorize

                              (x²-1)/(x-1)   =   (x-1)(x+1)/(x-1)

                                                  =    x+1

    Now applying the limit as x tends to 1, then we will get the answer 

                                                =  1+1 =2


2. Evaluate the limit as x approaches 2 of (x²+x-6)/(x²-2x).

Solution:

          In this problem also if we apply the limit directly, then the whole problem will become as 0/0. So let us factorize the numerator and denominator.

         (x²+x-6)/(x²-2x)      =     [(x+3)(x-2)]/ x(x-2)

                                        =        (x+3)/ x

            Substituting the value of x as 2,  we get the value as 5/2.

Type 3:

Solving limit problems by rationalizing the numerator:

1.  Evaluate the limit :

Solution:

                       In this problem we have to rationalize the numerator by multiplying the conjugate of the numerator.

                               =   [x - √(3x+4)][x+√(3x+4)]/(4-x)[x+√(3x+4)]

                               =  x² -(3x+4)/(4-x)[x+√(3x+4)]

                               =  x²-3x -4 / (4-x)[x+√(3x+4)]

                 Now factorize the numerator

                              =  (x-4)(x+1)/(4-x)[x+√(3x+4)]

                              =    (x+1)/ {-[x+√(3x+4)]}

                Now applying the limit as x tends to 4, we get the value as

                              =  (4+1)/{-[[4+√(3(4)+4))]}

                             =        -5/8.


      Students can try to solve the problem on their own, they can verify the steps with the steps discussed in 'Limits-examples' . If you are having any doubt you can contact us through mail, we will help you to clear your doubts.

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