# Slove this limit please ?

Mar 18, 2017

$= \frac{1}{5}$

#### Explanation:

We use well known trig limit:

${\lim}_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$

Or, specifically, it's inverse:

${\lim}_{\theta \to 0} \frac{\theta}{\sin \theta} = 1$

So:

${\lim}_{x \to 0} \frac{5 {x}^{2}}{{\sin}^{2} 5 x}$

$\frac{1}{5} {\lim}_{x \to 0} \frac{5 x}{\sin 5 x} \cdot \frac{5 x}{\sin 5 x}$

And by the product rule for limits:

$= \frac{1}{5} {\lim}_{x \to 0} \frac{5 x}{\sin 5 x} \cdot {\lim}_{x \to 0} \frac{5 x}{\sin 5 x}$

If we say that $\theta = 5 x$, we have:

$= \frac{1}{5} {\lim}_{\theta \to 0} \frac{\theta}{\sin \theta} \cdot {\lim}_{x \to 0} \frac{\theta}{\sin \theta}$

$= \frac{1}{5} \cdot 1 \cdot 1 = \frac{1}{5}$

Mar 18, 2017

$\frac{1}{5.}$

#### Explanation:

Recall that, ${\lim}_{\theta \to 0} \frac{\theta}{\sin} \theta = 1.$

Now, the Reqd. Lim.=${\lim}_{x \to 0} \frac{5 {x}^{2}}{{\sin}^{2} \left(5 x\right)}$

$= {\lim}_{x \to 0} 5 {\left\{\frac{x}{\sin} \left(5 x\right)\right\}}^{2}$

$= {\lim}_{x \to 0} 5 {\left\{\frac{5 x}{\sin} \left(5 x\right) \cdot \frac{1}{5}\right\}}^{2.}$

$= 5 {\left(\frac{1}{5}\right)}^{2} {\left\{{\lim}_{\left(5 x\right) \to 0} \frac{5 x}{\sin} \left(5 x\right)\right\}}^{2.}$

$= \frac{1}{5} {\left(1\right)}^{2.}$

$= \frac{1}{5.}$

Enjoy Maths.!