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Suppose that functions $f$ and $g$ are differentiable at $x=a$ and that $f(a)=g(a)=0 .$ If $g^{\prime}(a) \neq 0,$ show that$$\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{f^{\prime}(a)}{g^{\prime}(a)}$$without using L'Hôpital's rule. [Hint: Divide the numerator and denominator of $f(x) / g(x)$ by $x-a$ and use the definitions for $\left.f^{\prime}(a) \text { and } g^{\prime}(a) .\right$]

$$\begin{aligned}&\lim _{x \rightarrow a} \frac{f(x)}{g(x)}=\frac{f^{\prime}(a)}{g^{\prime}(a)}\\&\text { Hence Proved }\end{aligned}$$

Calculus 1 / AB

Chapter 3

TOPICS IN DIFFERENTIATION

Section 6

L Hopital's Rule; Indeterminate Forms

Functions

Limits

Derivatives

Differentiation

Continuous Functions

Applications of the Derivative

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University of Michigan - Ann Arbor

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okay, we have the limit. Is X goes to a of f of X divided by G of X. Now this could be re written because half of a is equal to zero. You can run it like this because G of A is equal to zero. We can write it like this. Now let's go ahead and we're going to divide the numerator and the denominator by X minus a X minus a X minus A. Now let's go ahead and split up the limits. Okay. And by definition of derivative, the numerator is equal to the derivative of F at a. And the denominator is equal to the derivative of G at a and that's it. That's what we're trying to show.

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