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I have been struggling to solve this limit.

What is the limit as $x$ approaches $45^0$ of $$\frac{\sqrt{2}\cos x-1}{\cot x-1}?$$

I know how to use L'Hospital's rule to calculate this limit and got the answer as $0.5$. But, how do I calculate the limit by manipulating the function? Please provide only some hints to proceed.

lsp
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Rajath Radhakrishnan
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    Never use degrees when doing calculus. – mrf Jan 09 '14 at 13:39
  • @mrf Is it like a convention or does it create problems if I use degrees? – Rajath Radhakrishnan Jan 09 '14 at 13:40
  • You want to know how to find the limit without using L'Hospital rule ? – lsp Jan 09 '14 at 13:44
  • There are lots of problems. For example, what is the derivative of $\sin x$ when you measure $x$ in degrees? – mrf Jan 09 '14 at 13:49
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    It is a bit excessive to say that one should never use degrees. One should recognize when using the angle variable in radians is essential to the method of calculation. In this problem, however, once the function is manipulated into a form where "direct substitution" can be applied, it doesn't matter whether 45º or $ \ \frac{\pi}{4} \ $ radians is used... – colormegone Jan 09 '14 at 14:01

5 Answers5

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multiply numerator and denominator by $\sqrt{2}\cos x+1$. you get $$\frac{2\cos^2 x-1}{(\cot x -1)( \sqrt{2}\cos x+1 )}$$ now $2\cos^2 x-1=\cos 2x$ and $\cos 2x=\frac{1-\tan^2x}{1+\tan^2 x}$

also $\cot x-1=\frac{1-\tan x}{\tan x}$ using all these you get $$\frac{\sqrt{2}\cos x-1}{\cot x-1}=\frac{\tan x(1+\tan x)}{(1+\tan^2 x)( \sqrt{2}\cos x+1 )}$$

Suraj M S
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Alternative method:

Your expression can be written as: $$ \lim_{x\to\pi/4} \dfrac{ \sqrt2 \cos x - 1} {\cos x - \sin x} \cdot\sin x $$ $$ = \lim_{h\to 0} \dfrac{ \sqrt2 \cos (h + \pi/4) - 1} {\cos (h + \pi/4) - \sin (h + \pi/4)} \cdot\sin (h + \pi/4) $$

(taking $ x = \pi/4 + h $. And hey, it just looks complicated here! It isn't. )

$$ = \lim_{h\to 0} \dfrac{ (\cos h - \sin h) - 1} {-2\sin h} \cdot\sin (h + \pi/4) $$ (simplifying using the $\sin(A+B), \cos (A+B)$ formulas)

$$ = \lim_{h\to 0} \dfrac{ 1 - \cos h + \sin h} {2\sin h} \cdot\sin (h + \pi/4) $$ $$ = \lim_{h\to 0} \;\;2\sin\left(\frac h 2 \right)\dfrac{ \sin(h/2) + \cos(h/2)} {2 \cdot 2\sin (h/2)\cos(h/2)} \cdot\sin (h + \pi/4) $$

Now, everything breaks down to $0.5$ nicely.

Parth Thakkar
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HINT

You have an expression which is of the type 0 / 0. Then, use L'Hospital and you will get your limit.

Claude Leibovici
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  • Sir, I know how to use L'Hospital's rule to calculate this limit and got the answer as 0.5. But, how do I calculate the limit by manipulating the function? – Rajath Radhakrishnan Jan 09 '14 at 13:39
  • @RajathKrishnaR. Your result is correct. The other solution is to build a Taylor expansion of the function around x = Pi/4. Limited to the first order, you will have 1/2 + 3 (x-Pi/4) /4 + ... If you need more clarification, or if this not what you expect, just let me know. – Claude Leibovici Jan 09 '14 at 13:44
  • But, evaluating limits using Taylor expansion is not there in out math school syllabus but these types of questions are asked for the exam. So, I think there is another method to evaluate this. Can you please tell another method sir? – Rajath Radhakrishnan Jan 09 '14 at 13:49
  • @RajathKrishnaR. Suraj M.S. gave you THE solution if you cannot use Taylor and you want to directly work the function. Cheers. – Claude Leibovici Jan 09 '14 at 13:52
  • Thank you sir for your interest in the question. – Rajath Radhakrishnan Jan 09 '14 at 13:56
  • @RajathKrishnaR. You are welcome. I am glad if I have been able to help you even just a little. Cheers. – Claude Leibovici Jan 09 '14 at 13:59
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You have $$ \lim_{x\to\pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1} $$ that you can transform using the substitution $t=x-\pi/4$, that is, $x=t+\pi/4$: \begin{align} \lim_{x\to\pi/4}\frac{\sqrt{2}\cos x-1}{\cot x-1}&= \lim_{t\to 0}\frac{\sqrt{2}(\frac{1}{\sqrt{2}}\cos t-\frac{1}{\sqrt{2}}\sin t)-1} {\frac{\cot t-1}{\cot t+1}-1}\\[2ex] &=\lim_{t\to 0}(\cos t-\sin t-1)\frac{\cot t+1}{-2}\\[2ex] &=-\frac{1}{2}\lim_{t\to 0}\frac{(\cos t-\sin t-1)(\cos t+\sin t)}{\sin t}\\[2ex] \end{align} You can surely go on from here.

egreg
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$$\lim_{x\to\frac\pi4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\sqrt2\lim_{x\to\frac\pi4}\frac{\cos x-\cos\frac\pi4}{\cot x-\cot\frac\pi4}$$

$$=\sqrt2\frac{\frac{d(\cos x)}{dx}_{(\text{ at }x=\frac\pi4)}}{\frac{d(\cot x)}{dx}_{(\text{ at }x=\frac\pi4)}}=\cdots$$

Alternatively,

$$F=\lim_{x\to\frac\pi4}\frac{\sqrt{2}\cos x-1}{\cot x-1} =\sqrt2\lim_{x\to\frac\pi4}\frac{\cos x-\cos\frac\pi4}{\cot x-\cot\frac\pi4}$$

$$=\sqrt2\lim_{x\to\frac\pi4}\frac{-2\sin\frac{x+\frac\pi4}2\sin\frac{x-\frac\pi4}2}{-\sin(x-\frac\pi4)}\cdot\sin x\sin\frac\pi4$$

$$=\sqrt2\lim_{x\to\frac\pi4}\frac{-2\sin\frac{x+\frac\pi4}2\sin\frac{x-\frac\pi4}2}{-2\sin\frac{x-\frac\pi4}2\cos\frac{x-\frac\pi4}2}\cdot\sin x\sin\frac\pi4$$

As $x\to\frac\pi4, x\ne\frac\pi4\implies \sin\frac{x-\frac\pi4}2\ne0 $

$$\implies F=\sqrt2\lim_{x\to\frac\pi4}\frac{\sin\frac{x+\frac\pi4}2}{\cos\frac{x-\frac\pi4}2}\cdot\sin x\sin\frac\pi4$$

$$=\sqrt2\frac{\sin\frac\pi4}{\cos0}\cdot\sin\frac\pi4\sin\frac\pi4=\cdots$$

lab bhattacharjee
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