Limit without using de l'hopital

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    l'hopital Limit
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To solve the limit lim (1+2^(1/x))/(3+2^(1/x)) as x approaches 0 without using L'Hôpital's rule, one must analyze the behavior of 2^(1/x) as x approaches 0 from both the positive and negative sides. As x approaches 0 from the positive side, 2^(1/x) approaches infinity, while from the negative side, it approaches 0. This leads to different limit results: for x→0+, the limit evaluates to 1, and for x→0-, it evaluates to 1/3. Additionally, to simplify (2x+3)/(x+1) as x approaches 0, the limit can be directly computed as 3. The discussion emphasizes the importance of understanding limits through case analysis rather than relying solely on L'Hôpital's rule.
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how can i solve lim (1+2^1/x)/(3+2^1/x) without using de l'hopital?
 
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sorry for x→0±
 
How would you simplify (2x+3)/(x+1)?
 
catapax said:
sorry for x→0±

Split it into cases. What lim x->0+ of 2^(1/x)? What about lim x->0-? You should be able to tell just by thinking about it.
 
I'm locking this post as a courtesy to Dick and haruspex, because the OP showed no effort.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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