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Establish the given limit formula

Prove the formula for the limit:

    \[  \lim_{x \to {\color{red}a}} \frac{\sin x - \sin a}{x - a} = \cos a. \]


    \[  \lim_{x \to {\color{red}a}} \frac{\sin x}{x} = 1. \]

( Note: I think there is an error in this problem as stated in Apostol. The statement in the book is for the limit as x \to 0. However, the limit as x \to 0 of this is -\frac{\sin a}{a} not \cos a (since it is a quotient of continuous functions and for a \neq 0 the denominator is nonzero at x = 0). I’ve changed the statement of the question to be the limit as x approaches a. This makes the given formula correct, so is probably what was intended.)

Proof. From Theorem 2.3 (g) (Apostol, p. 96) we know

    \[ \sin a - \sin b = 2 \sin \frac{a-b}{2} \cos \frac{a+b}{2}. \]

So, evaluating the limit we have,

    \begin{align*}  \lim_{x \to a} \frac{\sin x - \sin a}{x-a} &= \lim_{x \to a} \frac{2 \sin \frac{x-a}{2} \cos \frac{x+a}{2}}{x-a} \\  &= \lim_{x \to a} \left( \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} \cos \frac{x+a}{2} \right). \end{align*}

But then,

    \[ \lim_{x \to a} \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} = \lim_{y \to 0} \frac{\sin y}{y} = 1.\]


    \[ y = \frac{x-a}{2} \quad \text{ and } \quad y \to 0 \quad \text{ as } \quad x \to a. \]


    \begin{align*}  \lim_{x \to a} \frac{\sin \frac{x-a}{2}}{\frac{x-a}{2}} \cos \frac{x+a}{2} &= \lim_{x \to a} \cos \frac{x+a}{2} \\  &= \cos a. \qquad \blacksquare \end{align*}


  1. Anonymous says:

    Why are we allowed to change the variable of limits via substitution? I know it makes intuitive sense but given the rigorous treatment of everything else, this claim seems quite unsubstantiated

  2. Anonymous says:

    Thanks, I thought I was going crazy 🤪. By the way, the limit when x->0 is sin(a)/a, no minus sign. Cheers.

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