In this post, I’ll describe how to quickly calculate the following trigonometric limit. Once you learn this method, you’ll have a better idea of how L’Hospital’s Rule is used.
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $$
It turns out that \( \sin(0) = 0 \). So both the numerator, \( \sin(x) \), and the denominator, \( x \), are zero when \( x \) is. Hence, we can use L’Hospital Rule.
To use L’Hospital Rule, we use derivatives. The derivative of \( \sin(x) \) is \(\cos(x)\) and the derivative of \( x \) is \( 1 \). By L’Hospital’s Rule,
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \frac{\cos(x)}{1}. $$
It turns out that \( \cos(0) = 1 \). So,
$$ \lim_{x \to 0} \frac{\cos(x)}{1} = 1 \,. $$
Hence,
$$ \lim_{x \to 0} \frac{\sin(x)}{x} = 1 $$
which is the answer we’re looking for.
