# Evaluate the limit: lim_{x to 0} {sin 5x} / {sin7x}.

## Question:

Evaluate the limit:

{eq}\displaystyle \lim_{x \to 0} \dfrac {\sin 5x} {\sin7x} {/eq}.

## Limits:

The limit is the approach to some value of f(x) as x approaches to some value. The given limit can be evaluated by using the properties of limit. The following properties of limit are:

{eq}\displaystyle \lim_{x \to a}(f(x) \div g(x)) = \lim_{x \to a}f(x) \div \lim_{x \to a}g(x) \\ \displaystyle \lim_{x \to a} \alpha f(x) = \alpha\lim_{x \to a} f(x) {/eq}

## Answer and Explanation:

We have:

{eq}\displaystyle \lim_{x \to 0} \dfrac{\sin 5x}{\sin 7x} {/eq}

Multiplying the numerator and denominator by x, we get:

{eq}\displaystyle \lim_{x \to 0} \dfrac{x\sin 5x}{x\sin 7x} \\ \displaystyle \lim_{x \to 0} \dfrac{\sin 5x}{x} \cdot \dfrac{x}{\sin 7x} \\ \displaystyle \lim_{x \to 0} \dfrac{\sin 5x}{x} \div \dfrac{\sin 7x}{x} {/eq}

Using:

{eq}\displaystyle \lim_{x \to a}(f(x) \div g(x)) = \lim_{x \to a}f(x) \div \lim_{x \to a}g(x) {/eq}

We get:

{eq}\displaystyle \lim_{x \to 0} \dfrac{\sin 5x}{x} \div \lim_{x \to 0} \dfrac{\sin 7x}{x} \\ \displaystyle \lim_{x \to 0} \dfrac{5\sin 5x}{5x} \div \lim_{x \to 0} \dfrac{7\sin 7x}{7x} {/eq}

Using:

{eq}\displaystyle \lim_{x \to a} \alpha f(x) = \alpha\lim_{x \to a} f(x) {/eq}

We get:

{eq}\displaystyle 5\lim_{x \to 0} \dfrac{\sin 5x}{5x} \div 7\lim_{x \to 0} \dfrac{\sin 7x}{7x} {/eq}

Since,

{eq}\displaystyle \lim_{x \to 0} \dfrac{\sin x}{x} = 1 {/eq}

Implies that:

{eq}\displaystyle 5(1) \div 7(1) = \frac{5}{7} {/eq}

Thus,

{eq}\displaystyle \boxed{\lim_{x \to 0} \dfrac{\sin 5x}{\sin 7x} = \frac{5}{7}} {/eq}

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