AboutScotto Expertise Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.).
I also have answered some questions in
Physics (mass, momentum, falling bodies),
Chemistry (charge, reactions, symbols, molecules), and
Biology.
Experience Experience in the area: I have tutored students in all areas of mathematics for over 20 years.
Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors.
Awards and Honors: I have passed Actuarial tests 100, 110, and 135.
Publications Maybe not a publication, but I have respond to well oveer 3000 questions on the PC.
That's around 2,000 in basic math and 1,000 in advanced math.
Education/Credentials I aquired well over 40 hours of upper division courses. This was well over the number that were required.
I graduated with honors in both my BS and MS degree from Oregon State University.
I was allowed to jump into a few junior level courses my sophomore year.
Awards and Honors I have been nominated as the expert of the month several times.
All of my scores right now are at least a 9.8 average (out of 10).
Past/Present Clients My past clients have been students at OSU, students at the college in South Seattle,
referals from a company, friends and aquantenances, people from my church, and people like you.
Question How do you use L'Hospital's Rule to compute
1. Lim as x approaches pos 0 of (cos x - e^x)/(ln(1+x^2)
2. lim as x approaches 0 of (x/sin x)^(1/x^2)
Answer 1. Lim as x->+0 of (cos x - e^x)/(ln(1+x^2)
L'Hospital's Rule says to take the derivative of the numerator and the denominator, and then see what the quotient comes out to be.
The derivative of the numerator is -sinx - e^x.
The lim(x->0) (-sinx - e^x) is -sin(0) - e^0 = -0 - 1 = -1.
The derivative of the denominator is 2x/(1+x^2).
The lim(x->0)(2x/(1+x^2)) is 2*0/(1+0^2) = 0/1 = 0.
Using this, the quotient is -1/0 = -∞, so the
lim as x->+0 of (cos x - e^x)/(ln(1+x^2) = -1/0 = -∞.
For the numerator, lim(x->0)(ln(x/sinx)) = ln(lim(x->0)(x/sinx)).
It is known that the lim(x->0)(sinx/x) is 1, and we have the
lim(x->0)(x/sinx). This is the same as 1/lim(x->0)(sinx/x), so the answer is 1/1 = 1.
In case this needs to be shown, apply L'Hospital's Rule to the top.
The numerator is x and the denominator is sinx. The derivative of the numerator is 1 and the derivative of the denominator is cosx. As x -> 0, the numerator goes to 1 and the denominator goes to 1, so the limit is 1/1, which is 1.
For the denominator, the lim(x->0)(x²) = 0.
Now putting the answers to these limits back into the equation gives
e^(1/0) = ∞.
Humor: So if the first problem is -∞ and the second problem is ∞,
then the sum of the two might be 0. Does this mean this question is asking for nothing at all...
