Introduction for limit rules for calculus:
In mathematics, the concepts of a “limit” are used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concepts of limit allow one to, in the complete space; define a new point from a Cauchy sequence of previously defined points. Limit is essential to calculus (and mathematical analysis in general) and is used to define continuity, derivatives and integrals. (Source.Wikipedia)
Limit rules Calculus includes that differential calculus and integral calculus limits is used
Rules for Limit Calculus:
(1) If f(x) = k for all x, then`lim_(x->c)` f(x) = k.
(2) If f(x) = x for all x, then`lim_(x->c)` f(x) = c.
(3) If f and g are two functions possessing limits and k is a constant then
(i)`lim_(x->c)` k f(x) = k`lim_(x->c)` f(x)
(ii)`lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) +`lim_(x->c)` g(x)
(iii)`lim_(x->c)`[f(x) - g(x)] =`lim_(x->c)` f(x) -`lim_(x->c)` g(x)
(iv)`lim_(x->c)` [f(x) . g(x)] =`lim_(x->c)` f(x) `lim_(x->c)` g(x)
(v)`lim_(x->c)`` f(x)/g(x)` =`lim_(x->c)` f(x) / `lim_(x->c)`g(x), g(x) ? 0
(vi) If f(x) = g(x) then `lim_(x->c)`f(x) =`lim_(x->c)` g(x).
(4) `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)
Examples for Limit Rules Calculus:
Example 1:
Evaluate 7`lim_(x->1)` (x3 - 1) / (x- 1)
Solution:
7`lim_(x->1)`x3 - 1 / x- 1
=7* 3(1)3 - 1 [limit rules: `lim_(x->a)` `lim_(x->a)` xn - an / x - a = nan - 1 (a ? 0)]
= 7*3(1)2
= 21
Example 2:
Find 8`lim_(x->0)`{(1 + x)4 - 1}/x
Solution:
Put 1 + x = t so that t ? 1 as x ? 0 and x= t -1
8`lim_(x->0)`(1 + x)4 - 1/ x = 8`lim_(t->1)`(t4 - 1)4/ (t - 1) [limit rules: `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)]
= 8*4(1)4-1 = 4*8=32
Example 3:
calculate the value n so that`lim_(a->2)`an - 2n/a - 2= 32
Solution: We have
`lim_(a->2)`an - 2n/a - 2 = n2n - 1 [limit rules: `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)]
? n2n - 1 = 32 = 4 × 8 = 4 × 23 = 4 × 2 4 - 1
Comparing on both sides we get n = 4
Example 4:
3`lim_(x->0)`log `(1 + x)/x ` = 1
Solution:
We know that loge (1 + x) =`x/1` - `x^2/2` +`x^3/3` - …
loge `(1 + x) / x ` = 1 - `x/2 ` +`x^2/3` - …
Therefore 3`lim_(x->0)`loge` (1 + x)/x``lim_(x->0)`loge (`(1+0)/ 0` )
=3 loge 0
= 3*1=3.
Between, if you have problem on these topics Converting Decimals to Mixed Numbers, please browse expert math related websites for more help on How to Find the Percentage of something.
In mathematics, the concepts of a “limit” are used to describe the value that a function or sequence "approaches" as the input or index approaches some value. The concepts of limit allow one to, in the complete space; define a new point from a Cauchy sequence of previously defined points. Limit is essential to calculus (and mathematical analysis in general) and is used to define continuity, derivatives and integrals. (Source.Wikipedia)
Limit rules Calculus includes that differential calculus and integral calculus limits is used
Rules for Limit Calculus:
(1) If f(x) = k for all x, then`lim_(x->c)` f(x) = k.
(2) If f(x) = x for all x, then`lim_(x->c)` f(x) = c.
(3) If f and g are two functions possessing limits and k is a constant then
(i)`lim_(x->c)` k f(x) = k`lim_(x->c)` f(x)
(ii)`lim_(x->c)` [f(x) + g(x)] =`lim_(x->c)` f(x) +`lim_(x->c)` g(x)
(iii)`lim_(x->c)`[f(x) - g(x)] =`lim_(x->c)` f(x) -`lim_(x->c)` g(x)
(iv)`lim_(x->c)` [f(x) . g(x)] =`lim_(x->c)` f(x) `lim_(x->c)` g(x)
(v)`lim_(x->c)`` f(x)/g(x)` =`lim_(x->c)` f(x) / `lim_(x->c)`g(x), g(x) ? 0
(vi) If f(x) = g(x) then `lim_(x->c)`f(x) =`lim_(x->c)` g(x).
(4) `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)
Examples for Limit Rules Calculus:
Example 1:
Evaluate 7`lim_(x->1)` (x3 - 1) / (x- 1)
Solution:
7`lim_(x->1)`x3 - 1 / x- 1
=7* 3(1)3 - 1 [limit rules: `lim_(x->a)` `lim_(x->a)` xn - an / x - a = nan - 1 (a ? 0)]
= 7*3(1)2
= 21
Example 2:
Find 8`lim_(x->0)`{(1 + x)4 - 1}/x
Solution:
Put 1 + x = t so that t ? 1 as x ? 0 and x= t -1
8`lim_(x->0)`(1 + x)4 - 1/ x = 8`lim_(t->1)`(t4 - 1)4/ (t - 1) [limit rules: `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)]
= 8*4(1)4-1 = 4*8=32
Example 3:
calculate the value n so that`lim_(a->2)`an - 2n/a - 2= 32
Solution: We have
`lim_(a->2)`an - 2n/a - 2 = n2n - 1 [limit rules: `lim_(x->a)` xn - an/x - a = nan - 1 (a ? 0)]
? n2n - 1 = 32 = 4 × 8 = 4 × 23 = 4 × 2 4 - 1
Comparing on both sides we get n = 4
Example 4:
3`lim_(x->0)`log `(1 + x)/x ` = 1
Solution:
We know that loge (1 + x) =`x/1` - `x^2/2` +`x^3/3` - …
loge `(1 + x) / x ` = 1 - `x/2 ` +`x^2/3` - …
Therefore 3`lim_(x->0)`loge` (1 + x)/x``lim_(x->0)`loge (`(1+0)/ 0` )
=3 loge 0
= 3*1=3.
Between, if you have problem on these topics Converting Decimals to Mixed Numbers, please browse expert math related websites for more help on How to Find the Percentage of something.
