Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 17|

Evaluate

\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)sin\left(2x\right)}{1-cos\left(x\right)}\right)

SOLUTION:

Direct substitution of x=0 gives the indeterminate form \frac{0}{0}. Therefore, we should apply trigonometric identities. 

We know that sin\left(2x\right)=2sin\left(x\right)cos\left(x\right), so we can rewrite the original function as 

\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)sin\left(2x\right)}{1-cos\left(x\right)}\right)=\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)2\left(sin\left(x\right)cos\left(x\right)\right)}{1-cos\left(x\right)}\right)

=2\cdot \lim\limits_{x\to 0}\left(\frac{sin^2\left(x\right)cos\left(x\right)}{1-cos\left(x\right)}\right)

We also know the pythagorean identity, sin^2\left(x\right)=1-cos^2\left(x\right). So,

=2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-cos^2\left(x\right)\right)cos\left(x\right)}{1-cos\left(x\right)}\right)

=2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+cos\left(x\right)\right)\left(1-cos\left(x\right)\right)cos\left(x\right)}{1-cos\left(x\right)}\right)

=2\cdot \lim\limits_{x\to 0}\left(\left(1+cos\left(x\right)\right)cos\left(x\right)\right)

=2\cdot \left(\left(1+cos\left(0\right)\right)cos\left(0\right)\right)

=2\cdot \left(\left(1+1\right)\cdot 1\right)

=4

 

 

 

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