Q1: Let $f,g \colon [a,b] \rightarrow \mathbb{R}$ be differentiable functions. What is the correct formulation for the extended mean value theorem?

A1: There is $\hat{x}$ with $f^\prime(\hat{x}) = 0$.

A2: There is $\hat{x}$ with $f^\prime(\hat{x}) = f(a)$.

A3: There is $\hat{x}$ with $\frac{f^\prime(\hat{x})}{g^\prime(\hat{x})} = \frac{f(b) - f(a)}{g(b) - g(a)}$.

A4: There is $\hat{x}$ with $\frac{f^\prime(\hat{x})}{g^\prime(\hat{x})} = (f(b) - f(a) ) \cdot (g(b)-g(a))$.

Q2: Let $f,g \colon [-1,1] \rightarrow \mathbb{R}$ be differentiable functions with $ g^\prime(x) > 0$ for all $x$ and $\lim_{x \rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)}$ exists. What is a correct implication using l’Hospital’s rule?

A1: If $f(0) = g(0)$, then $ \displaystyle \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)} $.

A2: If $f(0) \neq g(0)$, then $ \displaystyle \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)} $.

A3: If $f(0) = g(0) = 0$, then $ \displaystyle \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)} $.

A4: If $f(0) = g(0) = 1$, then $ \displaystyle \lim_{x \rightarrow 0} \frac{f(x)}{g(x)} = \lim_{x \rightarrow 0} \frac{f^\prime(x)}{g^\prime(x)} $.

Q3: Let’s apply l’Hospital’s rule for the following limit for $a > 0$: $$ \lim_{x \rightarrow 0} \frac{ \log(1 + a x) }{ x } $$

A1: $\frac{1}{a}$

A2: $1$

A3: $0$

A4: $a$

A5: $2a$

Q4: Let’s apply l’Hospital’s rule for the following limit for $a > 0$: $$ \lim_{x \rightarrow 0} \frac{ \sin( a x) }{ a x } $$

A1: $\frac{1}{a}$

A2: $1$

A3: $0$

A4: $a$

A5: $2a$