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I am studying for an entrance exam and would like somebody to confirm my answer or point out mistakes I made. Answers are greatly appreciated!

Find a and b so that the following Limit exists. $$ L = \lim_{x\rightarrow 0} \frac{\cos^5{x} + ax + b}{x^2} $$

My solution approach was using l'Hôpital's rule so I set a to 0 and b to -1. -1 cancels the 1 from cos 0 so I get 0/0 then I can use l'Hôpital's rule. Having a = 0 I can use it again.

Is this approach right?

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    Sure. *First* set $b=-1$. Use L'Hospital's Rule. *Then* set $a=0$. End up with $\lim_{x\to 0}\frac{-5\cos^4 x\sin x}{2x}$. Limit of this *could* be done with L'Hospital's Rule, but shouldn't be, since $\lim_{x\to 0}\frac{\sin x}{x}=1$.2011-09-05
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    @André: You're missing a minus sign from the cosine derivative there.2011-09-05
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    @joriki: Thanks. Middle of the night.2011-09-05
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    @entrance_exam: I personally prefer your/André's approach to the ones in the answers. Those are good to know and can be useful in other circumstances, but they involve more details if you spell them out rigorously.2011-09-05
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    @joriki.. But the problem asks to find all pairs $(a,b)$ such that the limit exists and if you start saying: "ok let's put $b=-1$..", you are already considering a particular case. How would you proceed in order to eliminate all possible pairs except for $(a,b)=(0,-1)$?2011-09-05
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    The OP perhaps described the process non-optimally, in terms of a strategy based on L'H. Rule. It really should have been something like this. If $b \ne -1$, the thing blows up near $0$. So $b=-1$ is the only possibility worth chasing. Use L'H. Rule. In the expression we obtain, if $a \ne 0$, we get blow-up. So $a=0$ is the only thing worth chasing.2011-09-05

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