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Artusartos
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This is how L'Hopital's Rule is defined in our textbook:
"Let s signify a, a^+, a^-, \infty or - \infty and suppose f and g are differentiable functions for which hte following limit exists:
lim_{x \rightarrow s} \frac{f'(x)}{g'(x)} = L..................(1)
If
lim_{x \rightarrow s} f(x) = lim_{x \rightarrow s} g(x) = 0................(2)
or if
lim_{x \rightarrow s} |g(x)| = + \infty.....................(3)
then
lim_{x \rightarrow s} \frac{f(x)}{g(x)} = L....................(4)
Here is the part that I was having trouble with...
Note that the hypthesis (1) includes some implicit assumptions: f and g must be defined and differentiable "near" s and g'(x) must be nonzero "near" s. For example, if lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)} exists, then there g' is nonzero. The requirement that g' be nonzero is crucial; see Exercise 30.7.
This is what exercise 30.7 is:
This example is taken from [38] and is due to Otto Stolz, Math, annalen 15 (1879), 556-559. The requirement in Theorem 30.2 [the one that I wrote above] that g'(x) \not= 0 for x "near" s is important. In a careless application of L'Hopital's rule in which the zeros of g' "cancel" the zeros of f' erroneous results can be obtained. For x in R, let
f(x) = x + cosxsinx and g(x)= e^{sinx}(x+cosxsinx)
\frac{f'(x)}{g'(x)} = \frac{2e^{-sinx}cosx}{2cosx+f(x)} if cosx is not zero adn x>3.
Show that lim_{x \rightarrow \infty} \frac{2e^{-sinx}cosx}{2cosx+f(x)} = 0 and yet the limit lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} does not exist.
I'm a bit confused about this, because I don't think I understood that theorem correctly...so can anybody clarify it for me?
For example, with this exercise...when I graphed f(x) and g(x), they both looked continuous and smooth...so obviously they were differentiable. But I could not understand how g'(x) is zero "near" infinity, because (as I said), I'm having trouble understanding the theorem...
"Let s signify a, a^+, a^-, \infty or - \infty and suppose f and g are differentiable functions for which hte following limit exists:
lim_{x \rightarrow s} \frac{f'(x)}{g'(x)} = L..................(1)
If
lim_{x \rightarrow s} f(x) = lim_{x \rightarrow s} g(x) = 0................(2)
or if
lim_{x \rightarrow s} |g(x)| = + \infty.....................(3)
then
lim_{x \rightarrow s} \frac{f(x)}{g(x)} = L....................(4)
Here is the part that I was having trouble with...
Note that the hypthesis (1) includes some implicit assumptions: f and g must be defined and differentiable "near" s and g'(x) must be nonzero "near" s. For example, if lim_{x \rightarrow a^+} \frac{f'(x)}{g'(x)} exists, then there g' is nonzero. The requirement that g' be nonzero is crucial; see Exercise 30.7.
This is what exercise 30.7 is:
This example is taken from [38] and is due to Otto Stolz, Math, annalen 15 (1879), 556-559. The requirement in Theorem 30.2 [the one that I wrote above] that g'(x) \not= 0 for x "near" s is important. In a careless application of L'Hopital's rule in which the zeros of g' "cancel" the zeros of f' erroneous results can be obtained. For x in R, let
f(x) = x + cosxsinx and g(x)= e^{sinx}(x+cosxsinx)
\frac{f'(x)}{g'(x)} = \frac{2e^{-sinx}cosx}{2cosx+f(x)} if cosx is not zero adn x>3.
Show that lim_{x \rightarrow \infty} \frac{2e^{-sinx}cosx}{2cosx+f(x)} = 0 and yet the limit lim_{x \rightarrow \infty} \frac{f(x)}{g(x)} does not exist.
I'm a bit confused about this, because I don't think I understood that theorem correctly...so can anybody clarify it for me?
For example, with this exercise...when I graphed f(x) and g(x), they both looked continuous and smooth...so obviously they were differentiable. But I could not understand how g'(x) is zero "near" infinity, because (as I said), I'm having trouble understanding the theorem...
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