$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$
-$$\lim _ { x \rightarrow } \frac { \sin 5 x } { \sin 4 x }$$
A) $\frac { 4 } { 5 }$
B) does not exist
C) $\frac { 5 } { 4 }$
D) 0

Question

Quizplus · Accepted Answer

We can rewrite the given limit as 
$$\lim_{x \rightarrow 0} \frac{\sin(5x)}{5x} \cdot \frac{4x}{\sin(4x)} \cdot \frac{5}{4}$$
Using the given limit, we know that the first fraction approaches 1 as $x$ approaches 0. Using the fact that $\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1$, we can also see that $\lim_{x \rightarrow 0} \frac{\sin(4x)}{4x} = 1$. Therefore, the second fraction approaches 1 as $x$ approaches 0. Thus, the limit simplifies to 
$$\left(\lim_{x \rightarrow 0} \frac{\sin(5x)}{5x}\right) \cdot \left(\lim_{x \rightarrow 0} \frac{4x}{\sin(4x)}\right) \cdot \frac{5}{4} = 1 \cdot 1 \cdot \frac{5}{4} = \frac{5}{4}$$
So the answer is $\boxed{\textbf{(C)}}$.

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$\text { Find the limit using } \lim _ { x = 0 } \frac { \sin x } { x } = 1 \text {. }$