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Evaluate the following limits: (b) t 0 ( 3t)/(t) (c) x infinity ^-1((x-1)/(2x))

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2 parts
b
3
c

π3\frac{\pi}{3}

Step-by-step solution

5 steps

Part b· 2 steps

Evaluate lim⁡t→0sin⁡3tt\lim_{t \to 0} \frac{\sin 3t}{t}

  1. Recognize the standard limit form.

    The limit is of the form lim⁡x→0sin⁡axx=a\lim_{x \to 0} \frac{\sin ax}{x} = a.

  2. Apply the standard limit.

    In this case, a=3a=3.

    lim⁡t→0sin⁡3tt=3\lim_{t \to 0} \frac{\sin 3t}{t} = 3

    The final answer is 3\boxed{3}.

Part c

π3\frac{\pi}{3}

Evaluate lim⁡x→∞cos⁡−1(x−12x)\lim_{x \to \infty} \cos^{-1}\left(\frac{x-1}{2x}\right)

  1. Evaluate the limit of the argument inside the inverse cosine function.

    Let L=lim⁡x→∞x−12xL = \lim_{x \to \infty} \frac{x-1}{2x}.
    To evaluate this limit, divide both the numerator and the denominator by the highest power of xx, which is xx:

    L=lim⁡x→∞xx−1x2xxL = \lim_{x \to \infty} \frac{\frac{x}{x} - \frac{1}{x}}{\frac{2x}{x}} L=lim⁡x→∞1−1x2L = \lim_{x \to \infty} \frac{1 - \frac{1}{x}}{2}

    As x→∞x \to \infty, 1x→0\frac{1}{x} \to 0.

    L=1−02=12L = \frac{1 - 0}{2} = \frac{1}{2}
  2. Substitute the limit of the argument into the inverse cosine function.

    Since cos⁡−1(y)\cos^{-1}(y) is a continuous function for y∈[−1,1]y \in [-1, 1], we can write:

    lim⁡x→∞cos⁡−1(x−12x)=cos⁡−1(lim⁡x→∞x−12x)\lim_{x \to \infty} \cos^{-1}\left(\frac{x-1}{2x}\right) = \cos^{-1}\left(\lim_{x \to \infty} \frac{x-1}{2x}\right) =cos⁡−1(12)= \cos^{-1}\left(\frac{1}{2}\right)
  3. Find the value of cos⁡−1(12)\cos^{-1}\left(\frac{1}{2}\right).

    The value yy such that cos⁡y=12\cos y = \frac{1}{2} in the principal range [0,π][0, \pi] is π3\frac{\pi}{3}.

    cos⁡−1(12)=π3\cos^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{3}

    The final answer is π3\boxed{\frac{\pi}{3}}.

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