Calc Solutions: Day 4: L'Hospital/Special Limit Challenge Problem

Citation: "Calculus Early Transcendentals, 7th Edition", by James Stewart Section 3.4 The Chain Rule Problem 51: The figure shows a circular arc of length s and a choard of length d, both suspended by a central angle theta. Find the limit as theta approach 0 from the right of s/d:

Solution: The first task will be identifying s and d in terms of theta, since the limit is as theta approaches 0. Recall the formula for arc-length:

$Arclength=s=r\theta$

In order to rename d in terms of theta and r, we must employ trigonometry.

Notice that we can now relate d with theta and r via the sine function:

$sin(\frac{\theta}{2})=\frac{d}{2}/r$
Ergo,
$d=2rsin(\frac{\theta}{2})$
And this finally gives:

$\lim_{\theta\rightarrow 0^+}{\frac{r\theta}{2rsin{(\frac{\theta}{2})}}}=\lim_{\theta\rightarrow 0^+}\frac{\theta/2}{sin{(\frac{\theta}{2})}}$
There are two ways to complete this problem. One involves recalling a special limit, and the other involves L'Hospital's Rule. I will show each respectively:

Special Limit:
This route using the special sine limit:

$\lim_{x\rightarrow 0^+}{\frac{sin(x)}{x}}=1$
We can manipulate our limit to put it in this special form as follows:

$\lim_{\theta\rightarrow 0^+}{\frac{\theta/2}{sin(\theta/2)}}=\frac{1}{\lim_{\theta\rightarrow 0^+}{\frac{sin(\theta/2)}{\theta/2}}}$
By substituting x=theta/2 we can put this limit into a know form, particularly the special form sin(x)/x:

$\frac{1}{\lim_{\theta\rightarrow 0^+}{\frac{sin(\theta/2)}{\theta/2}}}=\frac{1}{\lim_{x\rightarrow 0^+}{\frac{sin(x)}{x}}}=\frac{1}{1}=1$
This completes the problem using the first method, which is what the section was asking for. However, for the more advanced students, and myself, I would initially approach the limit with the following method.

L'Hospital's Rule:

$\lim_{\theta\rightarrow 0^+}\frac{\theta/2}{sin{(\frac{\theta}{2})}}$
This limit has the misfortune of being 0/0 when direct substitution is applied. As you should remember (or note) this indicates a removable discontinuity. In this situation, we can also use L'Hospital's Rule which states:

Theorem: L'Hospital's Rule:
If

$\lim_{x \to c}f(x)=\lim_{x \to c}g(x)=0 \text{ or } \pm\infty$ , and

$\lim_{x\to c}\frac{f'(x)}{g'(x)}$ exists, and

$g'(x)\neq 0$ on an open interval containing c,

then

$\lim_{x\to c}\frac{f(x)}{g(x)} = \lim_{x\to c}\frac{f'(x)}{g'(x)}$ .

In our case, we have the following:
$f'(\theta)=1/2 \quad g'(\theta)=\frac{1}{2}cos(\theta)$
The conditions of L'Hospital's Rule (red) are certainly fulfilled, so we can apply the implication of the rule, as follows:

$\lim_{\theta\rightarrow\0^+}{\frac{\theta/2}{sin(\theta/2)}}=\lim_{\theta\rightarrow\0^+}{\frac{1/2}{1/2\,cos(\theta/2)}}=\lim_{\theta\rightarrow\0^+}{\frac{1}{cos(\theta/2)}}=\frac{1}{1}=1$

It's clear that both methods are valid approaches to the problem, yielding the same answer. From a personal standpoint, it's easier to remember the rule (L'Hospital's) that can be applied to numerous problems than to systematically manipulate equations so that the special (and important!) limit can be used

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Calc Solutions

Monday, January 23, 2012

Day 4: L'Hospital/Special Limit Challenge Problem

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