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If $ f' $ is continuous, $ f(2) = 0 $, and $ f'(2) = 7 $, evaluate$$ \displaystyle \lim_{x\to 0} \frac{f(2 + 3x) + f(2 + 5x)}{x} $$

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since $f(2)=0,$ the given limit has the form $\frac{0}{0}$\[\lim _{x \rightarrow 0} \frac{f(2+3 x)+f(2+5 x)}{x} \stackrel{\Perp}{=} \lim _{x \rightarrow 0} \frac{f^{\prime}(2+3 x) \cdot 3+f^{\prime}(2+5 x) \cdot 5}{1}=f^{\prime}(2) \cdot 3+f^{\prime}(2) \cdot 5=8 f^{\prime}(2)=8 \cdot 7=56\]

01:17

Wen Zheng

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Amrita Bhasin

Calculus 1 / AB

Calculus 2 / BC

Chapter 4

Applications of Differentiation

Section 4

Indeterminate Forms and l'Hospital's Rule

Derivatives

Differentiation

Volume

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Oregon State University

University of Nottingham

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for this problem, we are given a couple of givens. Um And that's that F prime is continuous. F of two equals zero, and F prime of two equals seven. So based on that, we can use direct substitution Trying to find the limit as X approaches zero. So once we do that, we have F of two Plus F of two. And that's going to give us zero plus 0/0 equals zero. So um we're gonna have the indeterminate form, which means we need to take the derivative. So when we take the derivative of both sides, We're going to have the limit as X approaches zero of 3. F crime of tube class three X plus five F. Prime. Uh Fighting crime of two plus five X. Um That's gonna ultimately boil down to three f prime of two plus five ethical crime of chill. And that we both know is going to be eight f prime of two since f prime of 27 That's going to give us 56. Our final answer.

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