Master L'Hospital's Rule: Solving Limits of (1/sin x) - (1/x)

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The discussion revolves around applying L'Hospital's Rule to solve the limit of the expression (1/sin x) - (1/x) as x approaches 0, which presents an indeterminate form of infinity minus infinity. The solution involves finding a common denominator and differentiating the numerator and denominator to evaluate the limit. After several steps, the limit is determined to be zero. Participants also clarify the behavior of the limit of 2cos(x)/sin(x) as x approaches 0, noting that it approaches infinity due to the properties of the tangent function. The conversation concludes with encouragement to tackle additional limit problems using similar techniques.
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we starting to learn about l'hospital's rule and i can't figure out this problem. It goes:

Find the Limit for the following indeterminate forms of the type "infinity - infinity"

limit as x approaches 0 of [(1/sin x)-(1/x)]
 
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God it's been too long and my memory of particular derivatives is fading, but if you find a common denominator of x sin(x) because:

\frac{1}{sin(x)}-\frac{1}{x} = \frac{x-sin(x)}{x \ sin(x)}

and we call the fraction

\frac{f(x)}{g(x)}

can you find the derivatives f'(x) and g'(x) to find the limit of the original form? You'll have to use the chain rule, right? :wink: Like I said it has been too long...hehe

UPDATE:
oops, that product rule, not chain rule!
 
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Okay ACLerok I couldn't help myself and gave it my best shot. See if you get the same thing:

let

\frac{x-sin(x)}{x \ sin(x)} = \frac{f(x)}{g(x)}

then

f'(x) = 1-cos(x)

and

g'(x) = sin(x) + x \ cos(x)

so the format is now

\lim_{x \rightarrow 0} \ \frac{1-cos(x)}{sin(x)+x \ cos(x)} = \frac{1-1}{0+(0)(1)}

that's indeterminate, too, so take the drivative again to get

\lim_{x \rightarrow 0} \ \frac{sin(x)}{2cos(x)-x \ sin(x)}

after some algebraic hoopla you get

\lim_{x \rightarrow 0} \ \frac{1}{\frac{2cos(x)}{sin(x)}-x} = \frac{1}{\infty - 0} = 0

So the limit is zero.

NOTE: I updated this page once to correct errors I made during the computation, starting with the SECOND derivative
 
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Hi and thanks a lot for replying. I understand everything you did but one thing. can you please explain why the limit of 2cos(x)/sin(x) as x approaches 0 equals infinity?
 
2cos(x)/sin(x) is 2/tan(x). tan(0) is 0, so 2cos(x)/sin(x) is 2/0.

cookiemonster
 
ACLerok said:
Hi and thanks a lot for replying. I understand everything you did but one thing. can you please explain why the limit of 2cos(x)/sin(x) as x approaches 0 equals infinity?
Cookiemonster said it all, but I wanted to add that an easy way to spot this sort of thing is punch sin(0.0000001) in the calc. You get a VERY small number, and if you punched a smaller number than 0.0000001 you'd get an even smaller number for the sin() value. Ad infinitum. If you divide by a fraction you multiply the numerator.

As long as the numerator is a positive value (no matter how big or how small) you get positive infinity for 1/sin(x) as x -> 0
 
Severian596 said:
... punch sin(0.0000001) in the calc. You get a VERY small number, ...
Make sure your calculator is in radians and you should see a moderately interesting result (which, incidently, supports the limit going to zero in the original difference expression).
 
cookiemonster said:
2cos(x)/sin(x) is 2/tan(x). tan(0) is 0, so 2cos(x)/sin(x) is 2/0.

cookiemonster

isn't 2/0 undefined?
 
Yes, but lim{x->0+} 2/x is infinity and lim{x->0-} 2/x is -infinity, so either way it goes to a big number.

Edit: I should point out that tan(x) is continuous in the neighborhood of 0, so that's why it goes to a big number.

cookiemonster
 
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  • #10
ACLerok said:
isn't 2/0 undefined?
Right. ACLerok keep in mind that when referring to limits as x -> 0 people tend to refer to fractions like

2/0

as

2/(an arbitrarily tiny number that's so close to zero the numerator is infinity)

Thanks for this thread, btw! It was fun to work through.
 
  • #11
ok thanks a lot guys! one down two more tough ones to go...
 
  • #12
http://www.eden.rutgers.edu/~cjjacob/images/limit.gif

is this problem exactly the same as the first one i posted or is there something i have to do first in order to differentiate?
 
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  • #13
Try putting the two terms over one denominator and simplifying.

cookiemonster
 
  • #14
Did you figure that last one out AC?
 
  • #15
no i actually just looked at a friend's answer.

this is the second problem i am given.

http://www.eden.rutgers.edu/~cjjacob/images/calcb.gif

after simplifying i get lim x->0, (1-x/x^2)

i figure that the lim x->0, (1/x^2), and x^2 being a very small number, is +infinity.

am i right or wrong?
 
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  • #16
ACLerok said:
no i actually just looked at a friend's answer.
I think you should not have given up so easily. The problem looks much worse than it actually is.

lim \ x \rightarrow 0 \ (\frac{1+x}{x}-\frac{1-x}{x})


= \frac{1+x-(1-x)}{x} = \frac{1+x-1+x}{x} = \frac{2x}{x} = 2
 
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  • #17
ACLerok said:
after simplifying i get lim x->0, (1-x/x^2)
Right
ACLerok said:
i figure that the lim x->0, (1/x^2), and x^2 being a very small number, is +infinity.
I like it
 
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