MHB Find Limit in 2 Mins - Tricks & Tips

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The discussion highlights a clever approach to finding the limit of the expression as x approaches infinity. It demonstrates that the limit can be transformed into a derivative form, specifically the derivative of e^x at x=0. The limit is calculated as 1, confirming that the derivative at that point is also 1. This method showcases the relationship between limits and derivatives effectively. Understanding this connection can simplify complex limit evaluations.
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We might recognize the shape of a derivative.
$$\lim_{x\to\infty} x(e^{\frac 1x}-1)
=\lim_{x\to 0} \frac 1x(e^x-1)
=\lim_{x\to 0} \frac{e^x-e^0}{x-0}$$

This is the derivative of $e^x$ at $x=0$, which is $e^0=1$.
 
Nice one!
 

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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