Rationale Behind t-Substitution for Evaluating Limits?

In summary, for the given limits involving radicals, t-substitutions can be determined by recognizing how the rules of algebra work and choosing a substitution that eliminates all roots in one step. This trick helps to turn the limits into ones that are easier to evaluate.
  • #1
bagasme
79
9
TL;DR Summary
How t-substitutions can be determined for any given limits involving radicals?
Hello all,

Given following limits:

  1. ##\lim_{x \rightarrow 1} {\frac {\sqrt x -1} {x^2 - 1}}##
  2. ##\lim_{x \rightarrow 1} {\frac {\sqrt {x+1} - 2} {x - 3}}##
  3. ##\lim_{x \rightarrow 1} {\frac {\sqrt[3] x - \sqrt[4] x} {\sqrt[6] x - \sqrt x}}##
Those limits can be evaluated by letting ##x = t^2##, ##x = t^2 - 1##, and ##x = t^{12}##, respectively for each limits.

I was curious, how such t-substitution (letting x with t) can be determined?

Bagas
 
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  • #2
This is a case of "Get rid of what disturbs most!" In these examples, it are the roots which disturb. So choosing a substitution which resolves the roots is a natural choice.

Do you understand, what such a substitution does geometrically?
 
  • #3
bagasme said:
Summary:: How t-substitutions can be determined for any given limits involving radicals?

Hello all,

Given following limits:

  1. ##\lim_{x \rightarrow 1} {\frac {\sqrt x -1} {x^2 - 1}}##
  2. ##\lim_{x \rightarrow 1} {\frac {\sqrt {x+1} - 2} {x - 3}}##
  3. ##\lim_{x \rightarrow 1} {\frac {\sqrt[3] x - \sqrt[4] x} {\sqrt[6] x - \sqrt x}}##
Those limits can be evaluated by letting ##x = t^2##, ##x = t^2 - 1##, and ##x = t^{12}##, respectively for each limits.

I was curious, how such t-substitution (letting x with t) can be determined?

Bagas

You don't need to do a substitution, just recognise how the rules of algebra work. For example:

##(x^2 - 1) = (x - 1)(x + 1) = (\sqrt x - 1)(\sqrt x + 1)(x + 1) \ \ ## (for ##x > 0##)
 
  • #4
fresh_42 said:
This is a case of "Get rid of what disturbs most!" In these examples, it are the roots which disturb. So choosing a substitution which resolves the roots is a natural choice.

Do you understand, what such a substitution does geometrically?
I don't know the geometry side.

Regarding t-substitution for formula 3), why had I substituted with LCM power (##t^{12}##)?
 
  • #5
bagasme said:
I don't know the geometry side.

Regarding t-substitution for formula 3), why had I substituted with LCM power (##t^{12}##)?
Because it eliminated all roots in one step.Geometrically it means that you run at a different pace towards the limit, that's all.
 
  • #6
It's a trick basically to turn the limits into ones you've seen before and can evaluate easily.
 

Related to Rationale Behind t-Substitution for Evaluating Limits?

1. What is t-substitution and when is it used to evaluate limits?

T-substitution, also known as u-substitution, is a technique used in calculus to simplify the evaluation of limits. It involves substituting a variable, typically denoted as t or u, into an expression to make it easier to evaluate. T-substitution is typically used when the limit involves a rational function or a trigonometric function.

2. How does t-substitution work?

To use t-substitution, we first identify a variable that will make the expression easier to evaluate. This variable is typically chosen to be the variable inside a function, such as x in f(x). We then substitute this variable with t, and rewrite the expression in terms of t. This allows us to evaluate the limit as t approaches a certain value, which can then be translated back to the original variable to find the limit.

3. What are the benefits of using t-substitution?

T-substitution can simplify the evaluation of limits by reducing the expression to a more manageable form. It can also help us to evaluate limits that may be otherwise impossible to solve, such as limits involving trigonometric functions or irrational expressions. Additionally, t-substitution can help us to identify indeterminate forms and determine the behavior of a function near a certain point.

4. Are there any limitations to using t-substitution?

While t-substitution can be a useful technique for evaluating limits, it may not always be the most efficient or appropriate method. In some cases, it may be easier to use other techniques such as L'Hôpital's rule or algebraic manipulation. Additionally, t-substitution may not work for all types of functions, such as those with multiple variables or complex expressions.

5. How can I practice and improve my skills in using t-substitution?

The best way to improve your skills in using t-substitution is to practice solving a variety of limit problems. You can find practice problems in calculus textbooks or online resources. It can also be helpful to review the steps and examples of t-substitution, and to seek guidance from a teacher or tutor if you are struggling with a specific problem.

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