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Please help me with this limit without using L'Hôpital's rule. I would by happy if you use simple solving. Thank you as much as I can ;).

$$\lim_{x\to0}{{\frac{\ln(\cos(-5x))}{\ln(\cos(-3x))}}}$$

Martin Sleziak
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Marián Slovák
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  • What are you did to edit my limit to this form? – Marián Slovák Nov 04 '15 at 22:23
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    This is kind of (very) similar to L'Hoptial: you can write $\ln(\cos(-5x)) = (-5x)\frac{\ln(\cos(-5x)) - \ln(\cos(-5\cdot 0)}{-5x}$ and interpret that as a derivative of some function when $x\to 0$. Another approach is to use the Taylor expansion of $\cos(x) = 1 - x^2/2 + O(x^4)$ and $\ln(1+x) = x + O(x^2)$ to deduce the limit. – Winther Nov 04 '15 at 22:25
  • (This question presents L'H approach, and then asks for other methods.) Bu the way, no need to drag those minuses around: cosine is an even function. –  Nov 04 '15 at 22:34

1 Answers1

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We have the following Taylor series approximations about zero ($v=0$ and $w=0$):

$$\ln(1+w)=w+O(w^2)\sim w$$

and

$$\cos(v)=1-\tfrac{1}{2}v^2+O(v^4)\sim 1-\tfrac{1}{2}v^2$$

Numerator: Put $v=-5x$ to get $\cos(-5x)\sim1-\tfrac{25}{2}x^2$ and then put $w=-\tfrac{25}{2}x^2$ to get $\ln(\cos(-5x))\sim-\tfrac{25}{2}x^2$.

Denominator: Put $v=-3x$ to get $\cos(-3x)\sim1-\tfrac{9}{2}x^2$ and then put $w=-\tfrac{9}{2}x^2$ to get $\ln(\cos(-3x))\sim-\tfrac{9}{2}x^2$.

Hence, the limit is

$$L=\lim_{x\to0}\frac{\tfrac{25}{2}x^2}{\tfrac{9}{2}x^2}=\frac{25}{9}$$

Marconius
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