## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 24

#### Answer

$2$

#### Work Step by Step

Let $\lim\limits_{t \to 0}f(t)=\lim\limits_{t \to 0}\dfrac{t-\sin t}{1-\cos t}$ But $f(0)=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ This implies that $\lim\limits_{t \to 0}\dfrac{(\sin t+t \cos t)}{\sin t}=\dfrac{0}{0}$ This shows that the limit has an indeterminate form, so we need to apply L-Hospital's rule as follows: $\lim\limits_{m \to n}f(x)=\lim\limits_{m \to n}\dfrac{p'(x)}{q'(x)}$ Thus, $\lim\limits_{t \to 0}\dfrac{2(\cos t)-(t) (\sin t)}{\cos t}=\dfrac{2 \cdot 1-(0)(0)}{1}=2$

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