Question

evaluate the.limit-​

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

Answer is m² / n²

We Have :-

$\lim_{ x \to 0} \frac{1-cosmx}{1-cosnx}$

Solution :-

There are 2 ways to do it :-

1 way :-

$\lim_{ x \to 0} \frac{1-cosmx}{1-cosnx} \\\\cosmx = 1-2sin^{2}\frac{mx}{2} \ [ cos2x = 1 - 2sin^{2} x ]\\\\cos nx = 1-2sin^{2}\frac{nx}{2} \\\\ \lim_{x \to 0} [ \frac{1-(1-2sin^{2}\frac{mx}{2}) }{1-(1-2sin^{2}\frac{nx}{2}) } ]\\\\ \lim_{x \to 0} \frac{2sin^{2}\frac{mx}{2} }{2sin^{2}\frac{nx}{2} } \\\\ \lim_{x \to 0} \frac{sin^{2}\frac{mx}{2} }{sin^{2}\frac{nx}{2} }$

$\lim_{x \to 0} \frac{(\frac{m}{2})^{2}[\frac{sin^{2}\frac{mx}{2} }{(\frac{m}{2} )^{2}}]}{(\frac{n}{2})^{2} [ \frac{sin^{2}\frac{nx}{2} }{(\frac{n}{2})^{2} } ] } \\\\\frac{m^{2}}{n^{2}} \frac{ \lim_{x \to 0} [ \frac{sin^{2}\frac{mx}{2} }{(\frac{m}{2})^{2} }] }{ \lim_{x \to 0} [ \frac{sin^{2}\frac{nx}{2} }{(\frac{n}{2})^{2} }] }$

$\frac{m^{2}}{n^{2}} * \frac{1}{1} \ \ \ [ \lim_{x \to 0} \frac{sinx}{x} =1 ] \\\\\frac{m^{2}}{n^{2}}$

2 way :-

$\lim_{x \to 0} ( \frac{1-cosmx}{1-cosnx})\\ \\Applying\ l'Hopital\\\\ \lim_{x \to 0} \frac{0-(-sinmx*m)}{0-(-sinnx*n)} \\ \\ \lim_{x \to 0} \frac{msinmx}{nsinnx} \\\\Again\ Using\ l'Hopital\\\\ \lim_{x \to 0} \frac{m^{2}cosmx}{n^{2}cosnx}$

$\frac{m^{2}}{n^{2}} \frac{ \lim_{x \to 0} cosmx }{ \lim_{x \to 0} cosnx } \\\\\frac{m^{2}}{n^{2}}$

Answer is m² / n²