In the mathematical expression the main concept of limit is used to express a value that a sequence or function approaches as the input or key approaches of some value. The limit is typically reduced as lim as in Lim(xn) = x or represent by the right arrow (→) as in an → a. Let us consider this function f(x) = x2. Examine that as x take values very close to 0, the value of f(x) also move towards 0. We say limits f(x) = 0 x →0
Rules For how to solve limits
Rule1: In learning online limits, given limits function put x=a .If f(a) is a definite value then
limits f(x) = f(a)
x →a
Rule2: In learning online limits, If proving limits f(x) is a rational function then factorize the numerator and the denominator.Cancel out the common factors and then put x=a
Rule3: If the given learning online limits function contains a surd then simplify it by using conjugate surd's.After simplification,put x =a
Rule4: If the given proving learning online limits function contains a series which is capable of being expanded then after making proper expansion and simplifying,cancel the common factors in the numerator and denominator,if any Then, put x =a
Limits Examples
1) Evaluate proving limits lim (xm -am ) / (xn -an)
x →a
Solution for proving limits: lim (xm -am /xn -an) = lim {xm -am /x-a) ÷ (xn -an /x-a)}
x →a x →a
Limits = lim (xm -am /x -a) ÷ lim(xn - an /x -a)
x →a x →a
Limits = (ma n-1) ÷ (nan-1)
Limits = ma m-1 / na n-1 = (m) /(n a m-n)
2) Evaluate proving limits lim (x+2)3/3 - (a +2)3/2 / x-a
x→a
Solution proving limits: lim (x+2)3/3 - (a +2)3/2 / x-a
x→a
(x +2)3/2 - (a+2)3/2
= lim ------------------------------------------
(x+2)→(a+2) (x +2) - (a +2)
------------------------------------------------------
= 3/2. (a+2)(3/2 -1) = 3/2(a +2)1/2 [ lim (xn -an /x -a) = nan-1]
x→a
3) Find Limit (x →2) {3x2-5x+7}
Solution:- Given Limit ( x →2) {3x2-5x+7}
= 3(2)2-5(2)+7 = 12-10+7 = 9
4) Show that Limit (x →3) (x2+2x-5) / (2x2-5x-1) = 5/2
Solution:-Limit (x →3) (x2+2x-5) / (2x2-5x-1)
= Limit (x →3) (x2+2x-5) / Limit ( x →3) (2x2-5x-1)
=[ (3)2+2(3)-5)] / [ 2(3)2-5(3)+1] = (9+ 9 - 5) / (18-15+1) = 10/ 4 = 5/2.
We can be solved these practice problems on limits by learning these limits problems.
Rules For how to solve limits
Rule1: In learning online limits, given limits function put x=a .If f(a) is a definite value then
limits f(x) = f(a)
x →a
Rule2: In learning online limits, If proving limits f(x) is a rational function then factorize the numerator and the denominator.Cancel out the common factors and then put x=a
Rule3: If the given learning online limits function contains a surd then simplify it by using conjugate surd's.After simplification,put x =a
Rule4: If the given proving learning online limits function contains a series which is capable of being expanded then after making proper expansion and simplifying,cancel the common factors in the numerator and denominator,if any Then, put x =a
Limits Examples
1) Evaluate proving limits lim (xm -am ) / (xn -an)
x →a
Solution for proving limits: lim (xm -am /xn -an) = lim {xm -am /x-a) ÷ (xn -an /x-a)}
x →a x →a
Limits = lim (xm -am /x -a) ÷ lim(xn - an /x -a)
x →a x →a
Limits = (ma n-1) ÷ (nan-1)
Limits = ma m-1 / na n-1 = (m) /(n a m-n)
2) Evaluate proving limits lim (x+2)3/3 - (a +2)3/2 / x-a
x→a
Solution proving limits: lim (x+2)3/3 - (a +2)3/2 / x-a
x→a
(x +2)3/2 - (a+2)3/2
= lim ------------------------------------------
(x+2)→(a+2) (x +2) - (a +2)
------------------------------------------------------
= 3/2. (a+2)(3/2 -1) = 3/2(a +2)1/2 [ lim (xn -an /x -a) = nan-1]
x→a
3) Find Limit (x →2) {3x2-5x+7}
Solution:- Given Limit ( x →2) {3x2-5x+7}
= 3(2)2-5(2)+7 = 12-10+7 = 9
4) Show that Limit (x →3) (x2+2x-5) / (2x2-5x-1) = 5/2
Solution:-Limit (x →3) (x2+2x-5) / (2x2-5x-1)
= Limit (x →3) (x2+2x-5) / Limit ( x →3) (2x2-5x-1)
=[ (3)2+2(3)-5)] / [ 2(3)2-5(3)+1] = (9+ 9 - 5) / (18-15+1) = 10/ 4 = 5/2.
We can be solved these practice problems on limits by learning these limits problems.
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