# How do you find lim sin^2x/x as x->0 using l'Hospital's Rule?

Sep 22, 2017

${\lim}_{x \to 0} \frac{{\sin}^{2} x}{x} = 0$

#### Explanation:

${\lim}_{x \to 0} \frac{{\sin}^{2} x}{x}$
If we apply limit then we get $\frac{0}{0}$ which is undefined.

Applying L'Hospital Rule
According to this rule we are going to differentiate numerator and denominator w.r.t x and then apply $\lim$.

now differentiate numerator and denominator w.r.t x
${\lim}_{x \to 0} \frac{2 \sin x \cos x}{1} = {\lim}_{x \to 0} \sin 2 x = \sin 2 \left(0\right) = 0$

Sep 22, 2017

${\lim}_{x \rightarrow 0} {\sin}^{2} \frac{x}{x} = 0$

#### Explanation:

We seek:

$L = {\lim}_{x \rightarrow 0} {\sin}^{2} \frac{x}{x}$

Both the numerator and the denominator $\rightarrow 0$ as $x \rightarrow 0$. thus the limit $L$ (if it exists) is of an indeterminate form $\frac{0}{0}$, and consequently, we can apply L'Hôpital's rule to get:

$L = {\lim}_{x \rightarrow 0} \frac{\frac{d}{\mathrm{dx}} {\sin}^{2} x}{\frac{d}{\mathrm{dx}} x}$

$\setminus \setminus = {\lim}_{x \rightarrow 0} \frac{2 \sin x \cos x}{1}$

Which we can now just evaluate to get:

$L = 2 \sin 0 \cos 0$
$\setminus \setminus = 2 \times 0 \times 1$
$\setminus \setminus = 0$

Note

This particular limit can also be readily evaluated without using L'Hôpital's rule as:

$L = {\lim}_{x \rightarrow 0} {\sin}^{2} \frac{x}{x}$
$\setminus \setminus = {\lim}_{x \rightarrow 0} \sin x \times \sin \frac{x}{x}$
$\setminus \setminus = {\lim}_{x \rightarrow 0} \sin x \times {\lim}_{x \rightarrow 0} \sin \frac{x}{x}$

The first limit can just be evaluated, and the second limit is a standard calculus result, and is equal to zero. So:

$L = \sin 0 \times 0$
$\setminus \setminus = 0$