Explanation A limit of a function at a point a is defined as the value that the function approaches as its argument approaches point a . The notation of limit is as follows: lim x → a ( f ( x ) ) = L , which is read as “limit of f ( x ) as x approaches a is L ” One-Sided Limits: A one-sided limit is a limit that only takes into account the values of a function that approaches a value from only one side i.e. either above or below. The right-side limit of a function f as it approaches a is give as lim x → a + ( f ( x ) ) = L The left-side limit of a function f as it approaches a is give as lim x → a − ( f ( x ) ) = L If right-side limit and left-side limit of a function are the same, then a two-sided limit exists for the function. Given: A limit as lim t → 0 + t c o s 1 t Formula Used: Squeeze Theorem – Let there be functions f, g, and h such that for all x ∈ [ a , b ] , then f ( x ) < h ( x ) < g ( x ) Also, suppose that, lim x → c f ( x ) = lim x → c g ( x ) = L , then for any a ≤ c ≤ b , lim x → c h ( x ) = L Calcul...

4th Edition

ISBN: 9781319279813

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 2, Problem 47CRE

To determine

The limit $\lim_{t \to 0^{+}} \sqrt{t} c o s \frac{1}{t}$ if the limit exists. If limit does not exist, then find out if one sided limit exists.

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Chapter 2, Problem 46CRE

Chapter 2, Problem 48CRE

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Chapter 2 Solutions

CALCULUS:EARLY TRANS.-SAPLINGPLUS

Ch. 2.1 - Prob. 1PQ Ch. 2.1 - Prob. 2PQ Ch. 2.1 - Prob. 3PQ Ch. 2.1 - Prob. 4PQ Ch. 2.1 - Prob. 1E Ch. 2.1 - Prob. 2E Ch. 2.1 - Prob. 3E Ch. 2.1 - Prob. 4E Ch. 2.1 - Prob. 5E Ch. 2.1 - Prob. 6E

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Knowledge Booster

The limit lim t → 0 + t c o s 1 t if the limit exists. If limit does not exist, then find out if one sided limit exists.

The limit limt→0+tcos1t if the limit exists. If limit does not exist, then find out if one sided limit exists.

Want to see the full answer?

Chapter 2 Solutions

The limit $\lim_{t \to 0^{+}} \sqrt{t} c o s \frac{1}{t}$ if the limit exists. If limit does not exist, then find out if one sided limit exists.