# Calculating the Limit $\lim\limits_{x\rightarrow 0} \frac{\cos^5{x} + ax + b}{x^2}$

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 $\lim_{x\to 0}\frac{\sin x}{x}=1$. – André Nicolas Sep 5 '11 at 9:34
@André: You're missing a minus sign from the cosine derivative there. – joriki Sep 5 '11 at 9:37
@joriki: Thanks. Middle of the night. – André Nicolas Sep 5 '11 at 9:38
@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 $(a,b)=(0,-1)$? – uforoboa Sep 5 '11 at 10:09
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. – André Nicolas Sep 5 '11 at 10:24

Your approach is correct. Since $\lim_{x\rightarrow 0}\left( \cos ^{5}x+ax+b\right) =1+b$, for the limit $$L=\lim_{x\rightarrow 0}\frac{\cos ^{5}x+ax+b}{x^{2}}$$ to exist, $1+b$ must be $0$, which means $b=-1$. And since $$\lim_{x\rightarrow 0}\left( \frac{d}{dx}\left( \cos ^{5}x+ax-1\right) \right) =a,$$ for L to exist, $a$ must be $0$. The limit is

$$L=\lim_{x\rightarrow 0}\frac{\frac{d}{dx}\left( \cos ^{5}x-1\right) }{\frac{d }{dx}\left( x^{2}\right) }=\lim_{x\rightarrow 0}-\frac{5}{2}\left( \cos ^{4}x\right) \frac{\sin x}{x}=-\frac{5}{2}.$$

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If you can use the Maclaurin series for $\cos(x)=1-\frac{1}{2}x^2+O(x^4)$, then try using the binomial theorem to get the first two non-zero terms for $\cos^5(x)$.

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@Thijs: It would seem that for an entrance exam, one should be able to use whatever they know. So I guess I should have said, "If you know the Maclaurin series..." – robjohn Sep 5 '11 at 13:48

Your approach is right, however in an exam i would reason as follows. The problem in your limit is the $x^2$ at the denominator. As long as you can factor it out from the fraction everything is allright. So try to write $$\cos(x)=1-\frac{x^2}{2}+o(x^2)$$ and you can easily see that $$\cos^5(x)=1+p(x)$$ where $p(x)$ is an infinite converging sum of monomials of degree at least $2$. Having noticed this, there is no way for your limit to exists unless $a=0$ and $b=-1$. This part of the reasinonig shows that, if you want your limit to exists, then necessarily $a=0,\: b=-1$. On the other hand, if $a=0,\: b=-1$, then a simple evaluation shows that $$\lim_{x\to 0}\frac{\cos^5(x)-1}{x^2}=\lim_{x\to 0}-\frac{5\cos^4(x)\sin(x)}{2x}=\lim_{x\to 0}-\frac{5\cos^4(x)}{2}\cdot\frac{\sin(x)}{x}=-\frac{5}{2}.$$ Hope everything is clear.

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Robjohn was quicker... – uforoboa Sep 5 '11 at 9:54
Generally, "speed" should not play a role in accepting answers. Rather, one should strive to accept the answer that one thinks will help the most readers. For example, if one answer serves a fish on a platter, but another answer teaches one how to fish, then generally one should accept the latter (this remark is not specific to this thread - it's a general remark). – Bill Dubuque Sep 5 '11 at 16:44

HINT $\$ If as $\rm\ x\to 0:\:$ $\rm\ f(x)\to f_0,\ \ f{\:\:'}(x)\to f_1,\ \ f(x)/x^2\to f_2\$ for $\rm\:f_i \in\mathbb R\$ then $\rm\ f_0 = 0 = f_1\:.$

Proof $\$ If $\rm\ f_0\ne 0\$ then $\rm\:f(x)/x^2\to\: f_0/0^+ = \infty\not\in\mathbb R\:.\:$ Thus $\rm\:f_0 = 0\:.\:$ Similarly $\rm\:f_1 = 0\:,\:\:$ else

$$\rm f_1 \ne 0\:\ \ \Rightarrow\ \ f_2\: =\ \lim_{x\:\to\: 0}\ \frac{f(x)}{x^2}\ =\ \lim_{x\: \to\: 0}\ \frac{\frac{f(x)-f(0)}{x}}{x}\ \to\ \frac{f_1}{0}\ =\ \pm \infty\ \not\in\: \mathbb R$$ $\$
So for $\rm\ f(x)\: =\: cos^5(x)+a\:x+b\:,\ \ f(0) = 0\:\Rightarrow\: b=-1\:,\:$ and $\rm\ f{\:\:'}(x)\: =\: -5\ cos^4(x)\ sin(x)+a\$ hence $\rm\: f{\:\:'}(0) = 0\:\Rightarrow\:a = 0\:.$

REMARK $\$ If you know about Taylor series then it should be clear that the above amounts to computing a Taylor series approximant.

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Something without l'Hospital. It is clear that $b=-1$ as mentioned above.

$$L = \lim_{x\rightarrow 0} \frac{\cos^5{x} + ax -1}{x^2}=\lim_{x \to 0}\left( \frac{\cos^5 x-1}{x^2}+\frac{a}{x} \right)$$

But

$$\lim_{x \to 0} \frac{\cos^5 x-1}{x^2}=\lim_{x \to 0} \frac{\cos x-1}{x^2}(1+\cos x+...+\cos^4 x)=\frac{5}{4} \lim_{x \to 0} \frac{-2\sin^2 \frac{x}{2}}{\frac{x}{2}^2}=-\frac{5}{2}$$

Therefore, if $a \neq 0$ the limit does not exist.

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