Recent content by redsox5

your numerator is wrong, her's the oringal again lim x-> infin x[(x+1)ln(1+(1/x))-1]

but it's 1/0 can I still use l'hospitals'?

after the second time i use it i have 1/0

how many time do you have to use l'hospital's rule..it's getting really messy

wait..i the correct answer -1 since it equals infin/neg infin

so after you use l'hospital's rule the first time you get \frac{ln(1+\frac{1}{x})+(x+1)(\frac{1}{1+\frac{1}{x}})}{\frac{-1}{x^{2}}} and that's infin/0 ...right?

you mean in my use of the chain rule?..is this the right approach though?

yea, I'm not reallu sure about that. I'm not sure if it will be acceptable

ok think i may have a solution lim x-> infin \frac{x}{[(x+1)ln(1+(1/x))-1]^-1} then use l'hospital's rule and get \frac{1}{-[(x+1)ln(1+(1/x))-1]^-1 * [ln(1+(1/x)+(x+1)(1/1+(1/x))}

can i leave x on top and move [(x+1)ln(1+(1/x))-1] to the bottom by raising it to the -1 power? Then use l'hospital's rule

well that's the oringal problem Is it an infinity times infinity problem, but to use l'hospital's rule it has to be a fraction

Homework Statement lim_{x -> infin} x[(x+1)ln(1+(1/x))-1] The Attempt at a Solution Am I able to start this with the chain rule. I'm not sure how to start this.

If I have a small model car and use a spring scale to measure the force, my reading is .130 kg. How do convert that to Newtons? Is gravity relevant to this equation? I know that F = m x a the mass of the car is .387 kg is the accerleration the acceleration of the car or gravity?

correction: it took 1.3 amps to electroylze the water

can someone just let me know if I'm calculating the force the rigt way?