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Limit of Algebraic Expressions

Limit of Algebraic Expressions

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Limit of Ratios


 »  for defined polynomials, algebra of limits apply.


 »  For polynomials evaluating to `0/0` value
    →  factorize numerator and denominator

    →  cancel common factors

    →  organize to standard results

    →  apply limit to modified expressions

Limit of Ratios

plain and simple summary

nub

plain and simple summary

nub

dummy

When the function does not evaluate to indeterminate value or undefined large, the limits of the function will be equal to the value got by substitution.
This generalization is not applicable only when the function is given in piecewise form.

If limits of numerator and denominator are `0`, then to find limit
 •  the common factors are canceled; or
 •  factors are separated to other standard form of limits.
Then find limit using the modified expression.

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Finding limit of standard ratios evaluating to `0/0` is explained with examples.


Keep tapping on the content to continue learning.
Starting on learning "Limits of functions as ratios". ;; Finding limit of standard ratios evaluating to 0 by 0 is explained with examples.

How to find limits of function `f(x)=color(deepskyblue)(3x^2+5x-2)/color(coral)(x^2+x-2)` at `x=2`?

  • Apply the limit to numerator and denominator
  • check if the function evaluate to `0/0`

The answer is 'check if function evaluate to `0/0`'

limits of function `f(x)=color(deepskyblue)(3x^2+5x-2)/color(coral)(x^2+x-2)` at `x=2`

By Substitution :
`f(x)|_(x=2)`
`quad quad = color(deepskyblue)(3(2)^2+5xx2-2)/color(coral)((2)^2+2-2)`
`quad quad = color(deepskyblue)(12+10-2)/color(coral)(4+2-2)`
`quad quad = color(deepskyblue)(20)/color(coral)(4)`
`quad quad = 5`

For left-hand-limit and right-hand-limit substitute `x=2-delta` and `x=2+delta`. You are required to work out this once.
It is generalized that, the left-hand-limit and right-hand-limit need not be computed. A standard result is specified as follows.

How to find limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

  • Apply the limit to numerator and denominator
  • check if the function evaluate to `0/0`

The answer is 'check if the function evaluate to `0/0`'

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

By Substitution :
`f(x)|_(x=2)`
`quad quad = color(deepskyblue)(3(2)^2-5xx2-2)/color(coral)((2)^2-2-2)`
`quad quad = color(deepskyblue)(12-10-2)/color(coral)(4-2-2)`
`quad quad = color(deepskyblue)(0)/color(coral)(0)`

When the function evaluates to `0/0`, examine if the numerator and denominator has a common factor.

factorize the numerator and denominator
`color(deepskyblue)(3x^2-5x-2) = color(deepskyblue)((3x+1)(x-2))`
`color(coral)(x^2-x-2) = color(coral)((x+1)(x-2))`
Note that the common factor `x-2` cannot be canceled at `x=2`. The function remains indeterminate at `x=2`.

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

Knowing that the numerator and denominator have a common factor, use the function `f(x)= color(deepskyblue)((3x+1)(x-2))/color(coral)((x+1)(x-2))`

left-hand-limit :
`lim_(x->2-)f(x)`
`quad quad = color(deepskyblue)((3(2-delta)+1)(2-delta-2))/color(coral)((2-delta+1)(2-delta-2))`
`quad quad = color(deepskyblue)((7-delta)(-delta))/color(coral)((3-delta)(-delta))`
The `-delta` from numerator and denominator can be canceled, as `-delta` is not `0`.
`quad quad = color(deepskyblue)(7-delta)/color(coral)(3-delta)`
In this, compared to `7` or `3`, the value of `delta` is negligible and that is reflected by substituting `delta = 0`.
`quad quad = color(deepskyblue)(7)/color(coral)(3)`

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

Knowing that the numerator and denominator have a common factor, use the function `f(x)= color(deepskyblue)((3x+1)(x-2))/color(coral)((x+1)(x-2))`

right-hand-limit :
`lim_(x->2+)f(x)`
`quad quad = color(deepskyblue)((3(2+delta)+1)(2+delta-2))/color(coral)((2+delta+1)(2+delta-2))`
`quad quad = color(deepskyblue)((7+delta)(delta))/color(coral)((3+delta)(delta))`
The `delta` from numerator and denominator can be canceled, as `delta` is not `0`.
`quad quad = color(deepskyblue)(7+delta)/color(coral)(3+delta)`
In this, compared to `7` or `3`, the value of `delta` is negligible and that is reflected by substituting `delta = 0`.
`quad quad = color(deepskyblue)(7)/color(coral)(3)`

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

`f(x)|_(x=2) = 0/0`

`lim_(x->2-)f(x)= 7/3`

`lim_(x->2+)f(x)= 7/3`

The limits are equal. So the function is defined.

Revising the steps taken to solve the problem...
limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-x-2)` at `x=2`

We found by Substitution :
`f(x)|_(x=2)`
`quad quad = 0/0`

And then factorized the numerator and denominator
`color(deepskyblue)(3x^2-5x-2) = color(deepskyblue)((3x+1)(x-2))`
`color(coral)(x^2-x-2) = color(coral)((x+1)(x-2))`

Though `(x-2)` cannot be canceled when `x=2`, we have worked out the left-hand-limit and right-hand-limit results, and know that the factor `x-2` will cancel as `delta` when limit is calculated.

From this example, a standard result is specified as "if a function evaluate to `0/0` at an input value, then the expected value of the function can be calculated by canceling out common factors between numerator and denominator."

So without going through the lengthy process of substituting `x=a-delta` and `x=a+delta`, just cancel out the common factor and find the limit for the rest of the function.

How to find limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-4x+4)` at `x=2`?

  • Apply the limit to numerator and denominator
  • check if the function evaluates to 0/0

The answer is 'check if the function evaluates to `0/0`'

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-4x+4)` at `x=2`

By Substitution :
`f(x)|_(x=2)`
`quad quad = color(deepskyblue)(3(2)^2-5xx2-2)/color(coral)(2^2-4(2)+4)`
`quad quad = color(deepskyblue)(12-10-2)/color(coral)(4-8+4)`
`quad quad = 0/0`

When the function evaluates to `0/0`, examine if the numerator and denominator has common factors.
factorize the numerator and denominator
`color(deepskyblue)(3x^2-5x-2) = color(deepskyblue)((3x+1)(x-2))`
`color(coral)(x^2-4x+4) = color(coral)((x-2)(x-2))`

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-4x+4)` at `x=2`

Knowing that the numerator and denominator have a common factor, use the function `f(x)= color(deepskyblue)((3x+1)(x-2))/color(coral)((x-2)(x-2))`

left-hand-limit :
`lim_(x->2-)f(x)`
`quad quad = color(deepskyblue)((3(2-delta)+1)(2-delta-2))/color(coral)((2-delta-2)(2-delta-2))`
`quad quad = color(deepskyblue)((7-delta)(-delta))/color(coral)((-delta)(-delta))`
The `-delta` from numerator and denominator can be canceled, as `-delta` is not `0`.
`quad quad = color(deepskyblue)(7-delta)/color(coral)(-delta)`
In this, compared to `7`, the value of `delta` is negligible and that is reflected by substituting `delta = 0`.
`quad quad = color(deepskyblue)(-7)/color(coral)(delta)`
`quad quad = -oo`

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-4x+4)` at `x=2`

Knowing that the numerator and denominator have a common factor, use the function `f(x)= color(deepskyblue)((3x+1)(x-2))/color(coral)((x-2)(x-2))`

right-hand-limit :
`lim_(x->2-)f(x)`
`quad quad = color(deepskyblue)((3(2+delta)+1)(2+delta-2))/color(coral)((2+delta-2)(2+delta-2))`
`quad quad = color(deepskyblue)((7+delta)(delta))/color(coral)((delta)(delta))`
The `delta` from numerator and denominator can be canceled, as `delta` is not `0`.
`quad quad = color(deepskyblue)(7+delta)/color(coral)(delta)`
In this, compared to `7`, the value of `delta` is negligible and that is reflected by substituting `delta = 0`.
`quad quad = color(deepskyblue)(7)/color(coral)(delta)`
`quad quad = oo`

limits of function `f(x)=color(deepskyblue)(3x^2-5x-2)/color(coral)(x^2-4x+4)` at `x=2`

`f(x)|_(x=2) = 0/0`

`lim_(x->2-)f(x)= -oo`

`lim_(x->2+)f(x)= oo`

So the function is not defined.

comprehensive information for quick review

Jogger

comprehensive information for quick review

Jogger

dummy

Limit of Quotient of defined polynomials: For a function `f(x)` having a ratio of two polynomials, if `f(x)|_(x=a)` is defined, then
`f(x)|_(x=a)`
`quad quad = lim_(x->a-) f(x)`
`quad quad = lim_(x->a+) f(x)`

Limit of polynomials evaluating to Quotient of `0` : For a function `f(x)= color(deepskyblue)(g(x))/color(coral)(h(x))` having a ratio of two polynomials, if `f(x)|_(x=a)` is `0/0` then factorize `color(deepskyblue)(g(x))` and `color(coral)(h(x))` such that
`color(deepskyblue)(g(x) = (x-a)^k g_1(x)g_s(x))`
`color(coral)(h(x) = (x-a)^l h_1(x)h_s(x))`
Where the ratio `color(deepskyblue)(g_s(x))/color(coral)(h_s(s))` is a standard form for which limit is defined at `x=a`.

Find the limit of the function as
`lim_(x->a) f(x)`
`quad quad = lim_(x->a) color(deepskyblue)((x-a)^k g_1(x))/color(coral)((x-a)^l h_1(x)) `
`quad quad quad quad xx lim_(x->a)color(deepskyblue)(g_s(x))/color(coral)(h_s(x))`
Cancel out the factors `x-a` from numerator and denominator.
Then find limit at `x=a` using the simplified expression.



           

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How to find limits of function f of x 3 x squared + 5 x minus 2 divided by x squared + x minus 2 at x=2?
apply;limit;numerator;denominator
Apply the limit to numerator and denominator
check;if;function;evaluate
check if the function evaluate to 0 by 0
The answer is "check if the function evaluate to 0 by 0"
Finding limit by substitution, left hand limit, and right hand limits is explained.;; It is generalized that the left-hand-limit and right-hand-limit need not be computed. A standard result is specified as follows.
When the function does not evaluate to indeterminate value or undefined large, the limits of the function will be equal to the value got by substitution. ;; This generalization is not applicable only when the function is given in piecewise form.
Limit of Quotient of defined polynomials: For a function f of x having a ratio of two polynomials, if f of x at x =a is defined, then f of x at x=a ;; equals limit x tending to a. minus, f of x ;; equals limit x tending to a. plus, f of x.
How to find limits of function f of x = 3 x squared minus 5 x minus 2 divided by x squared minus x minus 2 at x=2?
apply;limit;numerator;denominator
Apply the limit to numerator and denominator
check;if;function;evaluate
check if the function evaluate to 0 by 0
The answer is "check if the function evaluate to 0 by 0"
Limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus x minus 2, at x=2 ;; By substitution : f of x at x=2 ;; equals 3 into 2 squared minus 5 into 2 minus 2, by 2 squared minus 2 minus 2;; equals 12 minus 10 minus 2, by 4 minus 2 minus 2 ;; equals 0 by 0 ;; When the function evaluates to 0/0 , examine if the numerator and denominator has a common factor. factorize the numerator and denominator. 3 x squared minus 5 x minus 2, = 3 x + 1 multiplied x minus 2;; x squared minus x minus 2, = x + 1 multiplied x minus 2;; Note that the common factor x minus 2 cannot be canceled at x=2. The function remains indeterminate at x=2.
Limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus x minus 2, at x=2 ;; Knowing that the numerator and denominator have a common factor, use the function f of x = 3 x + 1 into x minus 2, divided by, x + 1 multiplied x minus 2;; left hand limit;; limit x tending to 2 minus, f of x ;; equals 3 into 2 minus delta plus 1 into 2 minus delta minus 2, divided by , 2 minus delta +1 into 2 minus delta minus 2 ;; equals 7 minus delta into minus delta, divided by , 3 minus delta into minus delta ;; The minus delta from numerator and denominator can be canceled, as minus delta is not 0 ;; equals 7 minus delta, by, 3 minus delta ;; In this, compared to 7 and 3 the value of delta is negligible and that is reflected by substituting delta = 0 ;; equals 7 by 3
Limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus x minus 2, at x=2 ;; Knowing that the numerator and denominator have a common factor, use the function f of x = 3 x + 1 into x minus 2, divided by, x + 1 multiplied x minus 2;; right hand limit;; limit x tending to 2 +, f of x ;; equals 3 into 2 + delta plus 1 into 2 + delta minus 2, divided by , 2 + delta +1 into 2 + delta minus 2 ;; equals 7 + delta into delta, divided by , 3 + delta into delta ;; The delta from numerator and denominator can be canceled, as delta is not 0 ;; equals 7 + delta, by, 3 + delta ;; In this, compared to 7 and 3 the value of delta is negligible and that is reflected by substituting delta = 0 ;; equals 7 by 3
limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus x minus 2, at x=2 ;; f of x at x=2, = 0 by 0 ;; limit x tending to 2 minus, f of x = 7 by 3 ;; limit x tending to 2 + f of x = 7 by 3;; The limits are equal . So the function is defined.
Revising the steps taken to solve the problem ;; limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus x minus 2, at x=2 ;; We found by Substitution : f of x at x = 2 , = 0 by 0; And then factorized the numerator and denominator ;; 3 x squared minus 5 x minus 2, = 3 x + 1 multiplied x minus 2;; x squared minus x minus 2, = x + 1 multiplied x minus 2;; Though x minus 2 cannot be canceled when x=2, we have worked out the left-hand-limit and right-hand-limit results, and know that the factor x-2 will cancel as delta when limit is calculated.;; From this example, a standard result is specified as "if a function evaluate to 0 by 0 at an input value, then the expected value of the function can be calculated by canceling out common factors between numerator and denominator." ;; So without going through the lengthy process of substituting x=a minus delta and x=a+delta, just cancel out the common factor and find the limit for the rest of the function.
If limits of numerator and denominator are 0, then to find limit ;; the common factors are canceled, or ;; factors are separated to other standard form of limits. ;; Then find limit using the modified expression.
Limit of polynomials evaluating to Quotient of 0 : For a function f of x =, g of x by h of x having a ratio of two polynomials, if f of x at x = a. is 0 by 0, then factorize g of x and h of x such that ;; g of x , = , x minus a power k, g 1 of x, g s of x ;; h of x , = , x minus a power l, h 1 of x, h s of x ;; where the ratio g s of x and h s of x is a standard form for which limit is defined at x=a. ;; Find the limit of the function as ;; limit x tending to a. f of x ;; equals limit x tending to a x minus a. power k, g 1 of k, divided by, x minus a. power l, h 1 of k, multiplied, limit x tending to a, g s of x, by, h s of x ;; Cancel out the factors x minus a. from numerator and denominator. Then find limit at x = a. using the simplified expression.
How to find limits of function f of x 3 x squared minus 5 minus 2, divided by, x squared minus 4 x plus 4; at x = 2?
apply;limit;numerator;denominator
Apply the limit to numerator and denominator
check;function;evaluates;0
check if the function evaluates to 0 by 0
The answer is "check if the function evaluates to 0 by 0"
limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus 4 x minus 4, at x = 2 ;; By substitution: f of x, at x = 2;; equals 3 multiplied 2 squared minus 5 times 2 minus 2, divided by 2 squared minus 4 times 2 plus 4 ;; equals 12 minus 10 minus 2, divided by 4 minus 8 + 4 ;; equals 0 by 0;; When the function evaluates to 0 by 0, examine if the numerator and denominator has common factors.;; Factorize the numerator and denominator;; 3 x squared minus 5 x minus 2, = , 3x+1 multiplied x minus 2;; x squared minus 4 x +4 = x minus 2 multiplied x minus 2
limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus 4 x minus 4, at x = 2 ;; Knowing that the numerator and denominator have a common factor, use the function f of x = 3x+1 multiplied x minus 2, divided by, x minus 2 multiplied x minus 2 ;; Left hand limit: limit x tending to 2 minus, f of x ;; equals 3 multiplied 2 minus delta +1, multiplied 2 minus delta minus 2, divided by, 2 minus delta minus 2 multiplied 2 minus delta minus 2 ;; equals 7 minus delta multiplied minus delta, divided by, minus delta multiplied minus delta ;; The minus delta from numerator and denominator can be canceled, as minus delta is not 0;; IN this, compared to 7, the value of delta is negligible and that is reflected by substituting delta = 0;; equals minus 7 by delta;; equals minus infinity
limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus 4 x minus 4, at x = 2 ;; Knowing that the numerator and denominator have a common factor, use the function f of x = 3x+1 multiplied x minus 2, divided by, x minus 2 multiplied x minus 2 ;; Right hand limit: limit x tending to 2 +, f of x ;; equals 3 multiplied 2 + delta +1, multiplied 2 + delta minus 2, divided by, 2 + delta minus 2 multiplied 2 + delta minus 2 ;; equals 7 + delta multiplied delta, divided by, delta multiplied delta ;; The delta from numerator and denominator can be canceled, as delta is not 0;; In this, compared to 7, the value of delta is negligible and that is reflected by substituting delta = 0;; equals 7 by delta;; equals infinity
Limits of function f of x = 3 x squared minus 5 x minus 2, divided by, x squared minus 4 x minus 4, at x = 2 ;; f of x at x =2, = 0 by 0;; limit x tending to 2 minus, f of x, = minus infinity;; limit x tending to 2 plus, f of x , = infinity;; So the function is not defined.

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