Questions: Use the limit as x approaches 0 of sin(x)/x=1 and/or the limit as x approaches 0 of (cos(x)-1)/x=0 to evaluate the following limit. The limit as x approaches 0 of sin(14x)/tan(x)

Solution Steps

To evaluate the limit \(\lim _{x \rightarrow 0} \frac{\sin 14 x}{\tan x}\), we can use the known limits \(\lim _{x \rightarrow 0} \frac{\sin x}{x}=1\) and \(\lim _{x \rightarrow 0} \frac{\cos x-1}{x}=0\). First, express \(\tan x\) as \(\frac{\sin x}{\cos x}\). Then, rewrite the original limit in terms of \(\frac{\sin x}{x}\) and \(\frac{\cos x}{x}\) to simplify the expression. Finally, apply the known limits to evaluate the expression as \(x\) approaches 0.

We start with the limit we want to evaluate: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{\tan(x)} \] Using the identity \(\tan(x) = \frac{\sin(x)}{\cos(x)}\), we can rewrite the limit as: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{\frac{\sin(x)}{\cos(x)}} = \lim_{x \rightarrow 0} \frac{\sin(14x) \cdot \cos(x)}{\sin(x)} \]

We can separate the limit into two parts: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{x} \cdot \lim_{x \rightarrow 0} \frac{\cos(x)}{\sin(x)} \] Using the known limit \(\lim_{x \rightarrow 0} \frac{\sin(kx)}{x} = k\) for \(k = 14\), we have: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{x} = 14 \] And since \(\lim_{x \rightarrow 0} \frac{\cos(x)}{\sin(x)} = \frac{1}{0} = \infty\), we need to evaluate the limit more carefully.

We can express the limit as: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{\sin(x)} \cdot \lim_{x \rightarrow 0} \cos(x) \] The limit \(\lim_{x \rightarrow 0} \cos(x) = 1\). Therefore, we focus on: \[ \lim_{x \rightarrow 0} \frac{\sin(14x)}{\sin(x)} = \frac{14}{1} = 14 \]

Final Answer