Chapter 11: Q. 5 (page 901)

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

Short Answer

Expert verified

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

Step by step solution

Step 1. Given information

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

Step 2. Finding limit for limt→0 1/(3+t)−1/3t

$lim_{t \to 0} \frac{1 / (3 + t) - 1 / 3}{t} = - \frac{1}{3 (t + 3)}$

$lim_{t \to 0} \frac{1 / (3 + t) - 1 / 3}{t} = - \frac{1}{3 (0 + 3)}$

$lim_{t \to 0} \frac{1 / (3 + t) - 1 / 3}{t} = - \frac{1}{9}$

Step 3. Finding limit for limt→0 (3+t)2−9t

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

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

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

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding limit for limt→0 1/(3+t)−1/3t

Step 3. Finding limit for limt→0 (3+t)2−9t

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Finding limits: Find the given limits if they exist. If a limit does not exist, explain why.limt→0 1/(3+t)−1/3t,(3+t)2−9t

Short Answer

Step by step solution

Step 1. Given information

Step 2. Finding limit for limt→0 1/(3+t)−1/3t

Step 3. Finding limit for limt→0 (3+t)2−9t

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Pure Maths

Probability and Statistics

Mechanics Maths

Geometry

Logic and Functions

Study anywhere. Anytime. Across all devices.

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