#13 Measure your Calibre

Calculus Level 4

$\large \lim_{x \to 0} \frac{\sqrt{\cos x} - \sqrt[3]{\cos x}}{\sin^2x} = \frac{A}{B}$

The limit above holds true for coprime integers $A$ and $B$ , where $A$ is negative and $B$ is positive. Find $A+B$ .

Bonus : Evaluate this limit without applying L'Hôpital's rule .

The answer is 11.

1 solution

Sabhrant Sachan
Mar 10, 2017

$\begin{aligned} L & = \displaystyle\lim_{x \to 0 } \dfrac{(\cos{x})^{\frac12}-(\cos{x})^{\frac13}}{\sin^{2}{x}} \\ & = \displaystyle\lim_{x \to 0 } \dfrac{(\cos{x})^{\frac12}-(\cos{x})^{\frac13}}{1-\cos^{2}{x}} \\ & =\dfrac{1}{2} \displaystyle\lim_{x \to 0 } \dfrac{(\cos{x})^{\frac12}-(\cos{x})^{\frac13}}{1-\cos{x}} \hspace{8mm} \small\text{Take } \cos{x} = t^{6} \\ & = \dfrac{1}{2}\displaystyle\lim_{t \to 1 } \dfrac{t^3-t^2}{1-t^6} \\ & = -\dfrac{1}{2}\displaystyle\lim_{t \to 1 } \dfrac{t^2(t-1)}{(t+1)(t-1)(t^4+t^2+1)} \end{aligned}$

$\boxed{ L = -\dfrac{1}{12} }$

How did you get rid of the 1+cosx and where did the 1/2 come from on the 3rd step?

Ashish Sacheti - 4 years, 2 months ago

Using the property of limits

$\displaystyle \lim_{x \to a} \left[ f(x) \cdot g(x) \right] = \lim_{x \to a} f(x) \cdot \lim_{x \to a} g(x)$

We can simplify

$\begin{aligned} \displaystyle \lim_{x \to 0} \dfrac{ {(\cos x)}^{1/2} - {(\cos x)}^{1/3} }{ 1 - \cos^2 x } &= \lim_{x \to 0} \dfrac{ {(\cos x)}^{1/2} - {(\cos x)}^{1/3} }{ (1+\cos x) (1-\cos x) } \\ &= \left( \lim_{x \to 0} \dfrac{1}{1+\cos x} \right) \cdot \left( \lim_{x \to 0} \dfrac{ {(\cos x)}^{1/2} - {(\cos x)}^{1/3} }{ 1 - \cos x } \right) \\ &= \dfrac 12 \cdot \lim_{x \to 0} \dfrac{ {(\cos x)}^{1/2} - {(\cos x)}^{1/3} }{ 1 - \cos x } \end{aligned}$

Tapas Mazumdar - 4 years, 2 months ago

I think that should be 2x, But Didnt it affected the answer? I guess it was a Typo while he was coding.

Md Zuhair - 2 years, 7 months ago

#13 Measure your Calibre

Other problems: Check your Calibre

The answer is 11.

1 solution

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