High School Rationale Behind t-Substitution for Evaluating Limits?

[bagasme]

TL;DR

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

 

[fresh_42]

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?

 

[PeroK]

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 $x > 0$)

 

[bagasme]

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 ($t^{12}$)?

 

[fresh_42]

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.

 

[hunt_mat]

It's a trick basically to turn the limits into ones you've seen before and can evaluate easily.