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Thought-Process to Discover Knowledge

Welcome to nubtrek.

Books and other education websites provide "matter-of-fact" knowledge. Instead, nubtrek provides a thought-process to discover knowledge.

In each of the topic, the outline of the thought-process, for that topic, is provided for learners and educators.

Read in the blogs more about the unique learning experience at nubtrek.continue

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



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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`
  • 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.

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(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)`

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`
  • check if the function evaluate 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-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.

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(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.

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
  • 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.

                            
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