Question

hw 2: problem 12
(1 point)
evaluate the limit:
$lim_{t \to 0} \frac{\tan 7t}{t \sec t} = $
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Explanation:

Step1: Rewrite trigonometric functions

Recall $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. Substitute these into the limit:
$$\lim_{t \to 0} \frac{\tan 7t}{t \sec t} = \lim_{t \to 0} \frac{\frac{\sin 7t}{\cos 7t}}{t \cdot \frac{1}{\cos t}} = \lim_{t \to 0} \frac{\sin 7t \cdot \cos t}{t \cdot \cos 7t}$$

Step2: Split the limit into products

Separate the terms to use the standard limit $\lim_{x \to 0} \frac{\sin x}{x} = 1$:
$$\lim_{t \to 0} \frac{\sin 7t}{t} \cdot \lim_{t \to 0} \frac{\cos t}{\cos 7t}$$

Step3: Adjust the sine term

Rewrite $\frac{\sin 7t}{t}$ as $7 \cdot \frac{\sin 7t}{7t}$, then apply the standard limit:
$$7 \cdot \lim_{t \to 0} \frac{\sin 7t}{7t} \cdot \lim_{t \to 0} \frac{\cos t}{\cos 7t}$$
Let $x=7t$, as $t \to 0$, $x \to 0$, so $\lim_{t \to 0} \frac{\sin 7t}{7t} = 1$.

Step4: Evaluate the cosine limits

Substitute $t=0$ into the cosine terms:
$$\lim_{t \to 0} \cos t = \cos 0 = 1, \quad \lim_{t \to 0} \cos 7t = \cos 0 = 1$$

Step5: Calculate the final value

Multiply all the evaluated parts:
$$7 \cdot 1 \cdot \frac{1}{1} = 7$$

Answer:

$7$