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I am looking for simple trigonometric or algebraic manipulation so that this limit can be solved without using L'Hôpital's rule $$ \lim_{x\rightarrow 0} \frac{\sin x^2}{ \ln ( \cos x^2 \cos x + \sin x^2 \sin x)} = -2$$

link on wolframalpha. Thank you for help!!

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3 Answers

up vote 9 down vote accepted

It is good recall the following asymptotics. $$\cos(x^2) = 1 + \mathcal{O}(x^4)$$ $$\cos(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$ $$\sin(x^2) = x^2 + \mathcal{O}(x^6)$$ $$\sin(x) = x + \mathcal{O}(x^3)$$

Hence, we get that $$\cos(x^2) \cos(x) = \left( 1 + \mathcal{O}(x^4) \right) \left( 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4) \right) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^4)$$ $$\sin(x^2) \sin(x) = \left(x^2 + \mathcal{O}(x^6) \right) \left( x + \mathcal{O}(x^3) \right) = \mathcal{O}(x^3)$$

Hence, we get that $$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = 1 - \dfrac{x^2}{2!}+ \mathcal{O}(x^3)$$ Hence, $$\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x)) = \ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right)$$

Also, recall that $$\ln(1+t) = t + \mathcal{O}(t^2).$$

Hence, $$\ln \left(1 - x^2/2 + \mathcal{O}(x^3) \right) = -\dfrac{x^2}{2} + \mathcal{O}(x^3)$$ Hence, $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \dfrac{x^2 + \mathcal{O}(x^{6})}{-x^2/2 + \mathcal{O}(x^3)} = \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)}$$

Hence, $$\lim_{x \to 0} \dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))} = \lim_{x \to 0} \dfrac{-2 + \mathcal{O}(x^4)}{1 + \mathcal{O}(x)} = \dfrac{\lim_{x \to 0} \left(-2 + \mathcal{O}(x^4) \right)}{\lim_{x \to 0} \left(1 + \mathcal{O}(x) \right)} = -2$$

EDIT Below is a slightly different method.

Note that $$\cos(x^2) \cos(x) + \sin(x^2) \sin(x) = \cos(x^2 - x)$$ We can rewrite $$\dfrac{\sin(x^2)}{\ln(\cos(x^2) \cos(x) + \sin(x^2) \sin(x))}$$ as $$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{x^2}{\ln(1 - 2\sin^2((x^2 - x)/2))}$$ We used the identity that $\cos(\theta) = 1 - 2 \sin^2 \left( \theta/2\right)$. $$\dfrac{\sin(x^2)}{\ln(\cos(x^2 - x))} = \dfrac{\sin(x^2)}{x^2} \times \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} \times \dfrac{x^2}{-2\sin^2((x^2 - x)/2)}$$ Now we have $$\lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = 1$$ $$\lim_{x \to 0} \dfrac{-2\sin^2((x^2 - x)/2)}{\ln(1 - 2\sin^2((x^2 - x)/2))} = 1$$ $$\lim_{x \to 0}\dfrac{x^2}{-2\sin^2((x^2 - x)/2)} = -2$$ Putting these together again gives us $-2$.

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Thank you for your answer ... but is there other way ... i mean simpler way without expansion ... just some trigonometric manipulation so that we get $-2 x/ln(1+x)$ type and get it's value?? –  Santosh Linkha Jun 4 '12 at 20:50
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Recall the subtraction law $$\cos(s-t)=\cos s\cos t+\sin s\sin t.$$ That improves the denominator to $\cos(x-x^2)$. We want to take the logarithm of that. It is nice to express the cosine as $\sqrt{1-\sin^2(x-x^2)}$. Take the logarithm, which is $(1/2)\ln(1-\sin^2(x-x^2))$. So we get a $2$ on top, and want to find the limit of $$\frac{\sin(x^2)}{\ln(1-\sin^2(x-x^2))}.$$ To finish, you will need to know something about the behaviour of $\frac{\ln(1-u)}{u}$ as $u$ approaches $0$ from the right. If you know that this limit is $-1$, then we can divide top and bottom by $\sin^2(x-x^2)$, and we are nearly finished. I can do the details if the finish is not obvious.

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By resorting to some elementary limits we immediately get the final answer. One of them is $\lim_{x\rightarrow 0} \ln(1+x)^\frac{1}{x}=1$ and the other one is $\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2}=-\frac{1}{2}.$ Therefore we have that:

$$ \lim_{x\rightarrow 0} \frac{\sin (x^2)}{ \ln ( \cos (x^2) \cos x + \sin (x^2) \sin x)} =\lim_{x\rightarrow 0} \frac{\sin (x^2)}{ \ln ( (\cos (x^2) \cos x)(1 + \tan (x^2) \tan x))}= \lim_{x\rightarrow 0} \frac{\sin(x^2)}{\ln{\cos(x^2)\cos(x)}}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\cos(x^2)\cos(x)-1}=\lim_{x\rightarrow 0} \frac{\frac{\sin(x^2)}{x^2}}{\frac{\cos(x)-1}{x^2}+\cos{x}\frac{\cos(x^2)-1}{x^2} }=\frac{1}{-\frac{1}{2}+0}=-2.$$

REMARK: $$\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2}=\lim_{x\rightarrow 0} \frac{\cos(x)-1}{x^2} \frac{\cos(x)+1}{\cos(x)+1} = \lim_{x\rightarrow 0} \frac{-\sin^2(x)}{2x^2}=-\frac{1}{2}$$ (no use of L'Hopital Rule)

Also notice that:

$$\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\ln{\cos(x^2)\cos(x)}}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{\frac{\ln(1+(\cos(x^2)\cos(x)-1))}{\cos(x^2)\cos(x)-1} (\cos(x^2)\cos(x)-1)}=\lim_{x\rightarrow 0} \frac{\sin(x^2)}{ \cos(x^2)\cos(x)-1}$$

Here we apply the trivial limit: $\lim_{u\rightarrow 0} \frac{\ln(1+u)}{u}=1$

The proof is complete.

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The replacing of $\cos x^2\cos x+\sin x^2\sin x$ in the log by $\cos x^2 \cos x$ is perhaps not obvious. –  André Nicolas Jun 4 '12 at 23:08
    
@André Nicolas: you're right. i've just improved that. –  Chris's sis Jun 4 '12 at 23:19
    
Can you explain about how you changed $\ln (\cos x \cos x^2 = (\cos x \cos x^2-1) $ –  Santosh Linkha Jun 5 '12 at 5:47
    
@experimentX: sure. I just added further explanations for it. –  Chris's sis Jun 5 '12 at 6:48
    
Oh ... thanks!! i wouldn't have understood that trick –  Santosh Linkha Jun 5 '12 at 6:50
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