If a, b satisfy lim[x -> 0] [(sqrt{ax + b} - 5) / x] = 1/2, then value of a + b is...?

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MethMath11

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i do not know how to solve it (My Country's Education Curriculum never teach me this type of question); I tried to differentiate it (dx and sorry for the bad grammar n english)

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No curriculum teaches how to solve every type of question; they give you tools by which you can attack unfamiliar problems! Don't expect to be spoon-fed. And don't blame the curriculum when you can't immediately see what to do. I don't either. But then I think ...

If you differentiated, that suggests you are trying to apply L'Hospital's rule. But you can only do that if the limit has the form 0/0. Under what conditions will that be true?
 
Dr.Peterson said:
No curriculum teaches how to solve every type of question; they give you tools by which you can attack unfamiliar problems! Don't expect to be spoon-fed. And don't blame the curriculum when you can't immediately see what to do. I don't either. But then I think ...

If you differentiated, that suggests you are trying to apply L'Hospital's rule. But you can only do that if the limit has the form 0/0. Under what conditions will that be true?
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well you gotta erase something
 
MethMath11 said:
i figure it out which one is numerator and denominator, thank you but i got a question, is it (1/2) x or 1/2x?
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It is (1/2)*x - (following not english but universal axiom of mathematics) translated to 1/2 times x - or 0.5 * x
 
MethMath11 said:
well you gotta erase something
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What does that mean?

Please answer my question. What must be true for this to have the form 0/0?

(I don't see where Khan is going with this; don't mix our recommendations together, because they may not be compatible.)
 
MethMath11 said:
i do not know how to solve it, tried to differentiate it (dx and sorry for the bad grammar n English)
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Dr.Peterson said:
Please answer my question. What must be true for this to have the form 0/0?
(I don't see where Khan is going with this; don't mix our recommendations together, because they may not be compatible.)
Click to expand...
Please do answer Prof. Peterson's question because you will need to use it.
I liked to play what I called the 'back-of-the-book game'.
We look up the answer and see: \(\displaystyle \bf{7.(C)}\). Now to play the game you have to explain why that is correct.
 
Dr.Peterson said:
...What must be true for this to have the form 0/0?...
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I also want to chime in to say this is exactly where I would begin as well. Answering this question will immediately give you \(b\), after which you may apply L'Hôpital's Rule to get \(a\). :)
 
Dr.Peterson said:
No curriculum teaches how to solve every type of question; they give you tools by which you can attack unfamiliar problems! Don't expect to be spoon-fed. And don't blame the curriculum when you can't immediately see what to do. I don't either. But then I think ...

If you differentiated, that suggests you are trying to apply L'Hospital's rule. But you can only do that if the limit has the form 0/0. Under what conditions will that be true?
Click to expand...
turn lim x - c f(x)/g(x) into
lim x - c f'(x)/g'(x) and i already tried it didnt give me the result that i want
 
MethMath11 said:
turn lim x - c f(x)/g(x) into
lim x - c f'(x)/g'(x) and i already tried it didnt give me the result that i want
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This gives me no new information. I already knew you were trying to apply L'Hopital's rule, and that it failed. I even told you why it failed (partially).

What we need from you is (a) the actual work you did, which may have value; and (b) an answer to my question. Have you considered whether the rule applies, and what that tells you?

Note in particular that if the numerator does not approach 0, then the whole expression has an infinite limit, not 1/2.
 
MethMath11 said:
i do not know how to solve it, tried to differentiate it (dx and sorry for the bad grammar n english)

View attachment 11585
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Putting an H over the equals, \(\displaystyle \mathop = \limits^H\), indicates the use l'Hospital's rule.
Now let \(\displaystyle b=25\) then \(\displaystyle \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {ax + 25} - 5}}{x}\mathop = \limits^H \mathop {\lim }\limits_{x \to 0} \frac{{\frac{a}{{2\sqrt {ax + 25} }}}}{1} = \mathop {\lim }\limits_{x \to 0} \frac{a}{{2\sqrt {ax + 25} }} = \frac{a}{2\cdot 5}\)
Now what is the value of \(\displaystyle a\) in order for the limit to be \(\displaystyle \tfrac{1}{2}~?\)
If you know then what is \(\displaystyle a+b~?\)
 
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