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This was my approach: $$\frac{\tan({\tan{x}})-\sin({\sin{x}})} {\tan{x}-\sin{x}}=\frac{\frac{\tan({\tan{x}})-\sin({\sin{x}})}{\tan{x}}} {\frac{\tan{x}-\sin{x}}{\tan{x}}}=\frac{\frac{\tan({\tan{x}})}{\tan{x}}-((\frac{\sin({\sin{x}})}{\sin{x}})\cdot \cos{x})} {1-\cos{x}}$$

$$\lim_{x \to 0} \frac{\frac{\tan({\tan{x}})}{\tan{x}}-((\frac{\sin({\sin{x}})}{\sin{x}})\cdot \cos{x})} {1-\cos{x}}=\frac{\lim_{x \to 0}\frac{\tan({\tan{x}})}{\tan{x}}-\lim_{x \to 0}\frac{\sin({\sin{x}})}{\sin{x}}\cdot{\lim_{x \to 0}(\cos{x})}}{1-\lim_{x \to 0}\cos{x}}=\frac{1-\lim_{x \to 0}\cos{x}}{1-\lim_{x \to 0}\cos{x}}=1$$ But the answer is 2.$$$$ For the method I got the answer 2 for, was where I used the the taylor series on the denominator and divided by $x^3$ and then expanded the numerator by taylor series again. I wanted to know what is wrong with my method. And where I have made the mistake.

q123LsaB
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    $\frac{1-\lim\cos x} {1-\lim\cos x}$ is a $\frac 00$ indeterminate form. You cannot cancel them. – insipidintegrator Jul 21 '22 at 06:15
  • isn't it the same as $~\lim{\frac{1-\cos{x}}{1-\cos{x}}}=> \frac{lim(1-\cos{x})}{lim(1-\cos{x})} =>\frac{1-lim cos{x}}{1-lim\cos{x}}~$. Here => means implies, and as for all the terms all the sums, and differences, and numerators and limits are computable. Isn't $~{1-\lim{\cos{x}} }~ $ close to 0? can't we take $~\lim{\cos{x}}~$=L and have$~\frac{1-lim cos{x}}{1-lim\cos{x}}=\frac{1-L}{1-L}$ – q123LsaB Jul 21 '22 at 06:29
  • See https://math.stackexchange.com/q/1659264/42969 or https://math.stackexchange.com/q/3376481/42969 for correct solutions. – Martin R Jul 21 '22 at 06:50
  • understood. I cant write Lim x/y = lim x/ limy if limy=0 – q123LsaB Jul 21 '22 at 07:06
  • i just wanted to know the mistake in my method, which i did. I do know the other methods, to solve the question. – q123LsaB Jul 21 '22 at 08:41
  • See math.stackexchange.com/a/3376505/42969, – Z Ahmed Jul 21 '22 at 10:29

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