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Establish the given limit formula

Prove the formula for the limit:

    \[ \lim_{x \to 0} \frac{\sin (5x)}{\sin x} = 5. \]

Using,

    \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]


Proof. First, since \lim_{x \to 0} \frac{\sin x}{x} = 1 we have

    \[ \lim_{x \to 0} \frac{\sin (5x)}{\sin x} = \lim_{x \to 0} \left( \left( \frac{\sin x}{x} \right) \left( \frac{\sin (5x)}{\sin x} \right) \right) = \lim_{x \to 0} \frac{\sin (5x)}{x}. \]

Then, we use the formula \sin (x+y) = \sin x \cos y + \cos x \sin y (from Apostol, Theorem 2.3 (f)), with 4x and x in place of x and y,

    \begin{align*}  \lim_{x \to 0}\frac{\sin (5x)}{x} &= \lim_{x \to 0} \frac{\sin (4x+x)}{x} \\  &= \lim_{x \to 0} \frac{\sin (4x) \cos x + \sin x \cos (4x)}{x} \\  &= \lim_{x \to 0} \left( \frac{2 \sin (2x) \cos (2x) \cos x}{x} + \frac{\sin x}{x} \cos (4x) \right) \\  &= \lim_{x \to 0} \left( \frac{\sin (2x)}{x} (2 \cos (2x) \cos x) + \frac{\sin x}{x} \cos (4x) \right) \end{align*}

But, from a previous exercise (Section 3.6, #15) we know \lim_{x \to 0} \frac{\sin (2x)}{x} = 2, so

    \begin{align*}  \lim_{x \to 0} \frac{\sin (5x)}{\sin x} = 2 \cdot 2 + 1 \cdot 1 = 5. \qquad \blacksquare \end{align*}

One comment

  1. SAUMYA says:

    sin(4x)+cos(x)+sinxcos(4x)/x. after this line iam not getting which formula you applied to proceed to next line will you please ellaborate with formula mentioned.

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