MHB How Can I Solve This Limit Without L'Hospital's Rule?

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The discussion focuses on solving a limit without using L'Hospital's Rule. Participants emphasize the importance of using fundamental limits for a more rigorous approach, specifically referencing the limit that leads to Euler's number, e. One user acknowledges forgetting the fundamental limit and corrects their solution to 2ln(10). There is a suggestion to evaluate the limit using L'Hospital's Rule for verification. The conversation highlights different methods for approaching limit problems in calculus.
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Hello MHB, how can i solve this limit without L'Hospital rule?
Here is a litle bit of my solution:
http://i.imgur.com/aEpCDY7.jpg
 
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Chipset3600 said:
Hello MHB, how can i solve this limit without L'Hospital rule?
Here is a litle bit of my solution:

Excellent!... of course l'Hopital's rule or Taylor expansion is more comfortable but the use of the 'fundamental limits' [in Your case $\displaystyle \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^{n} = e$...] is allwais more 'rigorous'...

Kind regards$\chi$ $\sigma$
 
chisigma said:
Excellent!... of course l'Hopital's rule or Taylor expansion is more comfortable but the use of the 'fundamental limits' [in Your case $\displaystyle \lim_{n \rightarrow \infty} (1 + \frac{1}{n})^{n} = e$...] is allwais more 'rigorous'...

Kind regards$\chi$ $\sigma$
Oh yeah! I forgot the fundamental limit :s, so is =2ln(10)...
Thanks
 
Chipset3600 said:
Oh yeah! I forgot the fundamental limit :s, so is =2ln(10)...
Thanks

Evaluate it using L'Hospital rule and verify your result .
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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