L' Hospital's Rule Square Root Problem

[Jason76]

(Sorry can't find the La-Tex symbol for Limits. Will try to look)

Find the limit as x approaches 1.

Step 1)

Plug in 1 into x.

\(\displaystyle \frac{x - 1}{\sqrt{x^{2} + 3} - 2}\)

This come out to being indeterminate or 0/0.

(Note: When factored out then it has a 0 in denominator, so that method doesn't work.)

So taking the rule:

Step 2).\(\displaystyle \frac{1}{\frac{x}{\sqrt{x^{2} + 3}}}\)

Step 3)\(\displaystyle \frac{\sqrt{x^{2}+ 3}}{x} \) Final Answer is 2.

Anyhow, I see the pattern of how the answer came about, but I don't understand the reasoning behind it. Well, I understand differintiation part. However. where did the fraction inside of the fraction come from in Step 2? Why did the square root in Step 2 get moved to the numerator in the final answer?

 

[mmm4444bot]

1/(b/a) = a/b

 

[Jason76]

Thanks for the help. Of course, another way to get the limit when the original problem is indeterminate is by factoring. In this particular case, you would need to "multiply by the conjugate". I think when it comes to "square roots" using the conjugate is much easier to handle.

 

[JeffM]

Thanks for the help. Of course, another way to get the limit when the original problem is indeterminate is by factoring. In this particular case, you would need to "multiply by the conjugate". I think when it comes to "square roots" using the conjugate is much easier to handle.

Yes, but they are trying to give you examples on how to use the new rule, not to teach a bunch of exceptions.

You asked for the reasoning behind the rule.

\(\displaystyle Given:\)

\(\displaystyle [1]: \displaystyle \lim_{x \rightarrow a} \phi(x) = 0 = \phi (a);\)

\(\displaystyle [2]: \phi'(a) = b.\)

\(\displaystyle [3]: \displaystyle \lim_{x \rightarrow a} \pi(x) = 0 = \pi (a) ;\)

\(\displaystyle [4]: \pi'(a) = c \ne 0;\ and\)

\(\displaystyle [5]: 0 < |h| << 1 \implies \pi(a + h) \ne 0.\)

\(\displaystyle So\ \phi (a + h) = \phi (a + h) - 0 = \phi(a + h) - \phi(a),\ and\ \pi(a + h) = \pi(a + h) - \pi (a).\)

With me so far?

\(\displaystyle So\ 0 < |h| << 1 \implies \dfrac{\phi(a + h)}{\pi(a + h)} = \dfrac{\phi(a + h) - \phi(a)}{\pi(a + h) - \pi(a)} = \dfrac{\dfrac{\phi(a + h) - \phi(a)}{h}}{\dfrac{\pi(a + h) - \pi(a)}{h}}.\)

\(\displaystyle But\ \displaystyle \lim_{h \rightarrow 0}\dfrac{\phi(a + h) - phi(a)}{h} = \phi'(a) = b.\)

\(\displaystyle And\ \displaystyle \lim_{h \rightarrow 0}\dfrac{\pi(a + h) - pi(a)}{h} = \pi'(a) = c \ne 0.\)

\(\displaystyle So \displaystyle \lim_{h \rightarrow 0}\left(\dfrac{\dfrac{\phi(a + h) - \phi(a)}{h}}{\dfrac{\pi(a + h) - \pi(a)}{h}}\right)\ exists\ and\ = \dfrac{b}{c}.\)

\(\displaystyle THUS\ \displaystyle \lim_{h \rightarrow 0}\left(\dfrac{\phi(a + h)}{\pi(a + h)}\right) = \lim_{x \rightarrow a}\dfrac{\phi(x)}{\pi(x)} = \dfrac{b}{c} = \dfrac{\phi'(a)}{\pi'(a)}.\)

PS Important edit. The proof above is NOT rigorous and does not deal with infinity.

Furthermore, the final conclusion above is not true if \(\displaystyle \phi(a)\) and \(\displaystyle \pi(a)\) are both real and both not zero.

if you look at the proof, you will see the importance of the functions being zero at a.

 

[girodd]

Of course, it is not necessarily the case that whoever posed the problem expects L'Hospital as a solution method. A rule of thumb is that if there is any other routine way to resolve an indeterminate form, you should use it in preference to L'H. Conjugation here is much to be preferred. L'H may have been good practice but it is also important to develop a taste for when it should be avoided.