cd246 said:
I would like to know if I did these the correct way.
Homework Statement
1.lim x->0 (1/sin 2x)-1/2x. the answer is 0.
2.lim x->0 (x^-5*ln x). The answer is -infinity.Homework Equations
1.I used L'Hopitals theorem
2.I derived them, than L'hopitals.The Attempt at a Solution
1.I only derived only the denominators for it to become (1/2cos2x)-(1/2), apply the 0 and it is (1/2)-(1/2)=0
Err... No, you cannot apply the rule like that. =.="
L'Hospital states that, if the limit \lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} is in one of the 2
Indeterminate Forms \frac{0}{0} ; \ \mbox{and } \frac{\infty}{\infty}, note that you can
only use L'Hospital for
(0/0), and
(inf/inf).
If the requirement above meets, then:
\lim_{x \rightarrow \alpha} \frac{f(x)}{g(x)} = \lim_{x \rightarrow \alpha} \frac{f'(x)}{g'(x)}. Note that, we differentiate
both numerator, and denominator.
You
cannot just differentiate the denominator, and leave the numerator unchanged. Well, that's not permitted. And you must have only
one fraction, f(x)/g(x).
So, what you should do is to change it into a single fraction, like this:
\lim_{x \rightarrow 0} \left( \frac{1}{\sin (2x)} - \frac{1}{2x} \right) = \lim_{x \rightarrow 0} \left( \frac{2x - \sin (2x)}{2x \sin (2x)} \right)
From here, there are 2 ways to do this, one is to use L'Hopital Rule, the other way is to use Taylor expansion for x around 0. They both work.
What Indeterminate Form is the expression above in? Is it 0/0, or inf/inf? Can you apply the rule now?
Can you go from here? :)
-------------------
Extra practice:
Find the limit of:
\lim_{x \rightarrow 0} \left( \frac{1}{\sin ^ 2 x} - \frac{1}{x ^ 2} \right)
2.f'(x) (-5x^-6)(1/x).
L'hopital's, f''(x) (30/x^7)or(30x^-7)
since the exponent is negative, is this the reason the answer is negative infinity?
No, no, no, again, you should read the L'Hospital Rule again, L'Hopital Rule only defined for
fraction, and
not multiplication. You should make it a fraction first, before applying the rule.
Or, you can do it slightly different. Say:
1. Is the above expression valid for x <= 0?
2. What are the limits of:
a. \lim_{x \rightarrow 0 ^ +} \ln (x)
b. \lim_{x \rightarrow 0 ^ +} x ^ {-5} = \lim_{x \rightarrow 0 ^ +} \frac{1}{x ^ 5}?
Can you go from here? :)
