Limit as x goes to 0 of ((sin(x)/x) - 1) /x

[newageanubis]

Homework Statement

Find the value of the limit as x goes to 0 of ((sin(x)/x) - 1) /x WITHOUT L'HOPITAL'S RULE.

Homework Equations

The Attempt at a Solution

I tried to get the expression inside the limit in terms of the fundamental trig limits, but all I got to was (1/x)*((sin(x)/x) -1). I want to say that the limit of this product as x -> 0 is equal to the desired derivative, and that since the limit of the ((sin(x)/x) -1) part as x -> 0 is 0, that the value of the limit is 0.

However, I'm not convinced that this solution follows from the basic limit laws. In fact, I don't think it does, since lim(1/x) as x-> 0 does not exist. How can I solve this question without using L'Hopital's Rule?

 

[Zondrina]

Homework Statement

Find the value of the limit as x goes to 0 of ((sin(x)/x) - 1) /x WITHOUT L'HOPITAL'S RULE.

Homework Equations

The Attempt at a Solution

I tried to get the expression inside the limit in terms of the fundamental trig limits, but all I got to was (1/x)*((sin(x)/x) -1). I want to say that the limit of this product as x -> 0 is equal to the desired derivative, and that since the limit of the ((sin(x)/x) -1) part as x -> 0 is 0, that the value of the limit is 0.

However, I'm not convinced that this solution follows from the basic limit laws. In fact, I don't think it does, since lim(1/x) as x-> 0 does not exist. How can I solve this question without using L'Hopital's Rule?

Have you heard of the squeeze theorem? Aka sandwich theorem. You can use this fact :

$|sin(x)| ≤ 1 \forall x \in ℝ$ ( Including (x) = (1/x) )

After massaging the absolute value of this particular (1/x)... remember you can take the limit of the entire inequality and bound your center function between the two outer ones.

 

[Dick]

The easy way would be to use the power series expansion of sin(x). Do you know that?