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junior college 2 | H3 Maths
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I'm stuck on this qn! can help on evaluating the limit for this pls?
Date Posted: 4 years ago
Views: 381
Both numerator and denominator would tend to 0, resulting in 0/0 which makes it undefined/indeterminate
So we use L'Hopital's rule here, where
lim [ f(x)/g(x) ] = lim [f'(x)/g'(x)]
You should be getting 16/27 as the limit if done correctly.
So we use L'Hopital's rule here, where
lim [ f(x)/g(x) ] = lim [f'(x)/g'(x)]
You should be getting 16/27 as the limit if done correctly.
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One way to do this is what we call the L Hopital Rule, which is applied when the limit form after substitution is like 0/0 or infinity/infinity.
In such situations, the rule lets us differentiate both the top and numerator in the hope that the new value obtained is able to be calculated.
In such situations, the rule lets us differentiate both the top and numerator in the hope that the new value obtained is able to be calculated.
Date Posted:
4 years ago
sorry is there any way to do this without the LHopital rule?
I did think of introducing the conjugate of the denominator at first and that was my first approach, but it led to a big mess.
If I manage to find a way, I will post a new one here.
EDIT: Found.
If I manage to find a way, I will post a new one here.
EDIT: Found.
Sent in the alternative approach already
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I found an alternative. We must do double conjugate. One for the numerator and one for the denominator.
I presume you are able to factorise the resulting denominator’s (x3 - 3x - 2) by Guess and check and Long division.
I presume you are able to factorise the resulting denominator’s (x3 - 3x - 2) by Guess and check and Long division.
Date Posted:
4 years ago
The x - 2 expression which I have found in the middle of my workings is the root cause of why the fraction becomes 0/0. Once this is eliminated, there will no longer be any x - 2 present in the denominator and all other expressions there cannot be zero, so we obtain a nice number.
Now I understand how to do this question. Thank you so much for your help!
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