Evaluate the following limit (2)

We know that

$\lim_{x\to 0}\frac{\sin x}{x} = 1$

So,

$\lim_{x\to 0}\frac{\sin{5x}}{\sin{4x}} = \frac54 \lim_{x\to 0}\frac{\frac{\sin{5x}}{5x}}{\frac{\sin{4x}}{4x}}$

$\lim_{x\to 0}\frac{\sin{5x}}{\sin{4x}} =\frac54 \frac{\lim_{x\to 0}\frac{\sin{5x}}{5x}}{\lim_{x\to 0}\frac{\sin{4x}}{4x}}$

$\color{#3D99F6}{\boxed{\lim_{x\to 0}\frac{\sin{5x}}{\sin{4x}} = \frac54 = 1.25}}$

Method 1: By L'Hôpital's rule

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin (5x)}{\sin (4x)} & \small \blue{\text{A 0/0 cases, L'Hôpital's rule applies}} \\ & = \lim_{x \to 0} \frac {5 \cos (5x)}{ 4\cos (4x)} & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = \frac 54 = \boxed{1.25} \end{aligned}$

Method 2: By Maclaurin series

$\begin{aligned} L & = \lim_{x \to 0} \frac {\sin (5x)}{\sin (4x)} & \small \blue{\text{By Maclaurin series}} \\ & = \lim_{x \to 0} \frac {5x - \frac {(5x)^3}{3!} + \frac {(5x)^5}{5!} - \cdots}{4x - \frac {(4x)^3}{3!} + \frac {(4x)^5}{5!} - \cdots} & \small \blue{\text{Divide up and down by }x} \\ & = \lim_{x \to 0} \frac {5 - \frac {5^3x^2}{3!} + \frac {5^5x^4}{5!} - \cdots}{4 - \frac {4^3x^2}{3!} + \frac {4^5x^4}{5!} - \cdots} \\ & = \frac 54 = \boxed{1.25} \end{aligned}$

• We first substitute $x=0$

$\large{\dfrac{\sin(5x)}{\sin(4x)}=\dfrac{0}{0}}$

• We get a $\dfrac{0}{0}$ situation, so we can apply L'Hôpital's rule.

• We differentiate both numerator and denominator to get:

$\large{\lim \limits_{x\rightarrow 0} \dfrac{\sin(5x)}{\sin(4x)} = \dfrac{\frac{d}{dx} (\sin(5x))}{\frac{d}{dx} (\sin(4x))}=\dfrac{ 5\cdot \cos(5x)}{4 \cdot \cos(4x)}}$

• Now we can substitute $x=0$ and get the answer.

$\large{\lim \limits_{x\rightarrow 0} \dfrac{\sin(5x)}{\sin(4x)} =\dfrac{5 \cdot \cos(0)}{4 \cdot \cos(0)}=\dfrac{5}{4}=\boxed{1.25}}$

$\lim\limits_{x\to 0} \frac{\sin(5x)}{\sin(4x)} \textrm{ is a } \frac{0}{0} \textrm{ situation so we can use L'Hopital's rule and differentiate the top and bottom}$

$\lim\limits_{x\to 0} \frac{\frac{d}{dx}(\sin(5x))}{\frac{d}{dx}(\sin(4x))} = \lim\limits_{x\to 0} \frac{5\cos(5x)}{4\cos(4x)} = \frac{5\cos(5(0))}{4\cos(4(0))} = \boxed{\frac{5}{4}=1.25}$

This may be unconventional, but this is how I did it:

Remove the $\sin$ function and $x$ from the numerator and denominator.

Answer: $\frac{5}{4}$ $= \fbox{1.25}$