Indeterminate mutterings

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Two young mathematicians consider a way to compute limits using derivatives.

Check out this dialogue between two calculus students (based on a true story):

Devyn: Riley! I think I have a problem!
Riley: What kind?
Devyn: I was calculating some limits last night.
Riley: Fun times.
Devyn: I realized something. L’Hôpital’s Rule lets us deal with $\zeroOverZero$ and $\inftyOverInfty$ forms.
Riley: Right!
Devyn: But what about all the other indeterminate forms? Like this one: $\displaystyle \lim _{x\to 0^+} \left ( \csc (x)(e^x-1)\right )$ .
Riley: So $\lim _{x\to 0^+} \csc (x) = \infty$ and $\lim _{x\to 0^+}e^x-1 = 0$ . This is a $\zeroTimesInfty$ -form, not $\zeroOverZero$ or $\inftyOverInfty$ .
Devyn: EXACTLY! L’Hôpital’s Rule only works on those fraction forms, not on this product form.
Riley: Can you take $\csc (x)(e^x-1)$ and write it like a fraction?
Devyn: Oh, like how $\csc (x) = \frac {1}{\sin (x)}$ ?
Riley: Yes, because then $\csc (x)(e^x-1) = \frac {e^x-1}{\sin (x)}$ .
Devyn: Great! That means my limit is the same as $\lim _{x\to 0^+} \frac {e^x-1}{\sin (x)}$ which is a $\zeroOverZero$ -form!

By L’Hôpital’s Rule, $\lim _{x\to 0^+}\frac {e^x-1}{\sin (x)} = \lim _{x\to 0^+} \frac {\answer {e^x}}{\answer {\cos (x)}} = \answer {1}$

Write the function $e^x$ as a fraction with a numerator of $1$ . $e^x = \frac {1}{\left (\answer {e^{-x}}\right )}$

Write the function $x+3$ as a fraction with a numerator of $1$ . $x+3 = \frac {1}{\left (\answer {\frac {1}{x+3}}\right )}$