L’Hôpital Rules the Graph

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In my last post on May 31, 2012 I showed a way of demonstrating why L’Hôpital’s Rule works. We looked at an example,

\displaystyle \underset{x\to \pi }{\mathop{\lim }}\,\frac{\tan \left( x \right)}{\sin \left( x \right)},

which met the requirement of the theorem called L’Hôpital’s Rule, namely the functions are differentiable and, since  tan(\pi ) = sin(\pi ) = 0 they intersect on the x-axis at \left( \pi ,0 \right). We looked at the graph and then zoomed-in at \left( \pi ,0 \right).

L’Hôpital’s Rule tells us that with these conditions the limit is the same as the limit of the ratio of their slopes (or their derivatives, if you prefer). Can you see what that ratio is from Figure 2? Even though this is not a “square window” the ratio is obviously –1.

Here are four other limits. See if you can find them by the method suggested here. Namely zoom-in on the point where the functions intersect and see if you can find the limits without doing any computations. (Yes, I know you already know the first 3, but try this idea anyway. )

1. \displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\sin \left( x \right)}{x}

2. \displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos \left( x \right)}{x}

3. \displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \left( \tfrac{\pi }{2}+x \right)}{x} also known as \displaystyle \underset{x\to 0}{\mathop{\lim }}\,\frac{\cos \left( \tfrac{\pi }{2}+x \right)-\cos \left( \tfrac{\pi }{2} \right)}{x}

4. \displaystyle \underset{x\to 1}{\mathop{\lim }}\,\frac{\pi \ln \left( x \right)}{\sin \left( \pi x \right)}

 

 

 

 

 

Answers in order:  1, 0, -1, -1

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