Chapter 11: Q. 5 (page 901)

Find the given limits if they exist. If a limit does not exist, explain why.
$\lim_{t \to 0} (\frac{\frac{1}{(3 + t) - \frac{1}{3}}}{t}, \frac{{(3 + t)}^{2} - 9}{t})$ .

Short Answer

Expert verified

The limit exists. The solution is,

$(- \frac{1}{3}, 6)$ .

Step by step solution

Step 1. Given Information.

The limit is,

$\lim_{t \to 0} (\frac{\frac{1}{(3 + t) - \frac{1}{3}}}{t}, \frac{{(3 + t)}^{2} - 9}{t})$ .

Step 2. Whether the limit exists or not.

Consider the limit,

$\lim_{t \to 0} (\frac{\frac{1}{(3 + t) - \frac{1}{3}}}{t}, \frac{{(3 + t)}^{2} - 9}{t})$

The objective is to evaluate the above limit.

$\lim_{t \to t_{0}} r (t) = \lim_{t \to t_{0}} (x (t), y (t), z (t)) = (\lim_{t \to t_{0}} x (t), \lim_{t \to t_{0}} y (t), \lim_{t \to t_{0}} z (t))$

Here, $\lim_{t \to t_{0}} x (t), \lim_{t \to t_{0}} y (t),$ and $\lim_{t \to t_{0}} z (t)$ exist.

Step 3. Finding the limit.

$\lim_{t \to 0} (\frac{1}{t} (\frac{1}{(3 + t)} - \frac{1}{3}), \frac{{(3 + t)}^{2} - 9}{t}) = \lim_{t \to 0} (\frac{1}{t} (\frac{3 - 3 - t}{(3 + t)}), \frac{9 + t^{2} + 6 t - 9}{t}) = \lim_{t \to 0} (\frac{1}{t} (\frac{- t}{3 + t}), \frac{t^{2} + 6 t}{t}) = \lim_{t \to 0} ((\frac{- 1}{3 + t}), t + 6) = (\lim_{t \to 0} (\frac{- 1}{3 + t}), \lim_{t \to 0} (t + 6)) = (- \frac{1}{3}, 6)$

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Find the given limits if they exist. If a limit does not exist, explain why.
$\lim_{t \to 0} (\frac{\frac{1}{(3 + t) - \frac{1}{3}}}{t}, \frac{{(3 + t)}^{2} - 9}{t})$ .

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Whether the limit exists or not.

Step 3. Finding the limit.

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Find the given limits if they exist. If a limit does not exist, explain why.limt→013+t-13t,3+t2-9t.

Short Answer

Step by step solution

Step 1. Given Information.

Step 2. Whether the limit exists or not.

Step 3. Finding the limit.

One App. One Place for Learning.

Most popular questions from this chapter

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Find the given limits if they exist. If a limit does not exist, explain why.
$\lim_{t \to 0} (\frac{\frac{1}{(3 + t) - \frac{1}{3}}}{t}, \frac{{(3 + t)}^{2} - 9}{t})$ .