# evaluate limit

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Hi, I need help with this question

is this right?

Thanks

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The top and bottom equaling zero screams L'Hospital.

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The top and bottom equaling zero screams L'Hospital.

I don't know L'Hospital. We have to solve these quesions using difference of cubes.

It is correct.

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umm i know im really late, but when you do use L'Hopital, not Hospital lol, hes not sick.., yea you get the same result.

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The top and bottom equaling zero screams L'Hospital.

The problem with L'Hospital's rule is that it requres one to know what derivatives are and as such what limits are. It can be the case that you have a circular argument, one needs the derivative that depends on the limit you are trying to prove etc...

So I would say be careful with L'Hospital's rule.

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Yes but thats when you want a rigid proof. If you wish for a quick solutions, then L'Hopital is the way to go, usually.

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do u wonder why its name l'hopital wen he didnt even invent it, and never claimed to..

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L'Hopital, not Hospital lol, hes not sick..

L'Hospital is correct in the sense that L'Hospital himself spelled his name with an 's' (but the 's' was silent).

L'Hôpital is the modern French spelling.

L'Hopital is least correct.

You like tomato and I like tomahto. L'Hospital, L'Hôpital let's call the whole thing off.

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Yes but thats when you want a rigid proof. If you wish for a quick solutions, then L'Hopital is the way to go, usually.

Absolutely correct.

In a "practical" situation I would use L'Hopital's rule, but for a rigid proof it may not be the best approach.

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Why wouldn't you always use L'Hopital's rule? As long as the functions are differentiable sufficiently often to obtain a conclusive result, this rule is valid, isn't it? And if so, then a result, obtained with this rule is equally proof as a result, obtained in another (usually more cumbersome) way.

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So what is wrong with L'Hospital's rule (or L'Hopital)?

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I think its only valid when direct substitution produces one of these indeterminate forms-

0/0,

or +- infinity/ +- infinity.

If it produces inderterminate forms such as-

1^(+- infinity)[which isnt still 1],

0^0 (which isnt 1 or 0),

or infinity^0, (which is not 1..)

then you can use logarithims to change it abit, then u can use the rule.

i dont think it works when the denominators derivative is 0 btw.

it is also potentially problematic, as certain functions as NOT, differentiable sufficiently often to obtain a conclusive result, as stated by woelen. They may oscillate wildly, and we may not notice. or, not just oscillation, but say, cusp, sudden change, like |x|. you could define 2 tangents, very problematic. L'hopitals rule is nice and easy to use in your school exams, but if your presenting a proof go the hard way, or at least substitute your example in the rules proof, makes it appear more rigours and easier to see if there are any cusps, oscillations etc

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i dont think it works when the denominators derivative is 0 btw.

The L'Hopital rule is not needed when the denominator is 0, while the numerator is non-zero. In that case the result is conclusive.

it is also potentially problematic, as certain functions as NOT, differentiable sufficiently often to obtain a conclusive result, as stated by woelen. They may oscillate wildly, and we may not notice. or, not just oscillation, but say, cusp, sudden change, like |x|. you could define 2 tangents, very problematic. L'hopitals rule is nice and easy to use in your school exams, but if your presenting a proof go the hard way, or at least substitute your example in the rules proof, makes it appear more rigours and easier to see if there are any cusps, oscillations etc

Whether a function is oscillatory or not does not tell anything about differentiability. Also, cusps and the like are not a problem. There are rigorous rules, which describe how many times a function is differentiable in a certain point. So, this reasoning is not valid here. I indeed see no problem to use the L'Hopital rule, also for very wild functions, as long as they are sufficienty often differentiable.

A function like |x| is not well-behaved in 0, but for all other values it is well-behaved. Try to read a text on real analysis, in order to obtain precise definitions of concepts like differentiability.

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For the first problem, I just multiplied the numerator and denominator by , (ax+1)^1/3 +1. Then I separated the terms and took the limit as x goes to zero. Not 100% sure if that's right.

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