Solving 2 Queries on 0/0 and x=(1/(1/x))

[DorumonSg]

2 questions...

Firstly, how can we tell if the quotient is 0/0, or infinite/infinite, I know that obviously we can sub the variable in as 0 and if the result of both denominator and numerator is 0, we can apply the L'Hospital Rule.

But how can we tell whether its a 0/0 type or infinite/infinite type? And can we apply the rule if its a 0/infinite type or infinite/0 type?

2nd questions...

I just found out x = (1/(1/x)) Is this correct? Why is that so? How do you poof it? I know that x = 1/(x^-1) so if we invert it again... its 1/((1/(x)^-1))^-1 but how do you get (1/(1/x))?

 

[mg0stisha]

1) You evaluate the limit as x approaches whatever number is given. If the top and bottom are both 0 or both infinite (along with another hand-full of indeterminate forms), you can then apply l'Hopital's rule.

2) x = (1/(1/x)) is true. try substituting x=2 into the equation. what's your answer? also, 1/(x^-1) = (1/(1/x)).

 

[DorumonSg]

1) You evaluate the limit as x approaches whatever number is given. If the top and bottom are both 0 or both infinite (along with another hand-full of indeterminate forms), you can then apply l'Hopital's rule.

2) x = (1/(1/x)) is true. try substituting x=2 into the equation. what's your answer? also, 1/(x^-1) = (1/(1/x)).

How bout 0/infinity or inifinity/0 can we use the Lhospital rule on that?

 

[lanedance]

How bout 0/infinity or inifinity/0 can we use the Lhospital rule on that?

a better question is do you need to...

you use L'Hops rule becaus the form is indeterminant

if your limit takes the form 1/infinity what does the limit become? so with that in mind what do you think 0/infinity tends o?