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Solving a limit by substitution

Hi, I'm having trouble computing this limit by a method described in my text book.

The limit I have to compute is

I can solve it by plugging in 1 to each of the x values or by the difference of squares and cancelling out so I end up with both of these have the result of 3

In my book the question says compute the following limits for by substituting

In my book it says: '...Another approach is to simply perform a change of variables by setting x := a +- δ where a is the value we want to evaluate the limit and δ is a number that is close to zero such that δ -> 0. This enables x to become as close as possible to a without actually being equal to a.


Can anyone help me understand this please?

Thankyou.
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davros
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Original post by 2902326
Hi, I'm having trouble computing this limit by a method described in my text book.

The limit I have to compute is

I can solve it by plugging in 1 to each of the x values or by the difference of squares and cancelling out so I end up with both of these have the result of 3

In my book the question says compute the following limits for by substituting

In my book it says: '...Another approach is to simply perform a change of variables by setting x := a +- δ where a is the value we want to evaluate the limit and δ is a number that is close to zero such that δ -> 0. This enables x to become as close as possible to a without actually being equal to a.


Can anyone help me understand this please?

Thankyou.


I think all they're suggesting is that for the example you've given, if you define

x=1+δx = 1 + \delta

then you now need to evaluate the limit as δ→0\delta \to 0 of an expression that depends on delta rather than x.

I don't think it makes things any easier in this case to be honest!

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