Lim x - 01-cos 5 x-1-cos x

Lim_x→ 01 cos5x/1 cos6x =lim_x→ 02sin^25x/2/lim_x→ 02sin^23x=2 (lim_x→ 0sin5x/2/5x/2 )^2×25/4x^2/2 ( lim_x→ 0sin3x/3x )^2× 9x^2 =2 × 1 ×25/4/2 × 1 × 9=25/4 × 9 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Standard Limits to Remove Indeterminate Form

lim x → 01-co...

Question

${lim}_{x \to 0} \frac{1 - c o s 5 x}{1 - c o s 6 x}$

Open in App

Solution

${lim}_{x \to 0} \frac{1 - c o s 5 x}{1 - c o s 6 x}$

$= \frac{{lim}_{x \to 0} 2 s i n^{2} \frac{5 x}{2}}{{lim}_{x \to 0} 2 s i n^{2} 3 x} = \frac{2 {⎛ ⎜ ⎝ {lim}_{x \to 0} \frac{s i n \frac{5 x}{2}}{\frac{5 x}{2}} ⎞ ⎟ ⎠}_{x \to 0}^{2} \times \frac{25}{4} x^{2}}{2 {({lim}_{x \to 0} \frac{s i n 3 x}{3 x})}_{x \to 0}^{2} \times 9 x^{2}}$

$= \frac{2 \times 1 \times \frac{25}{4}}{2 \times 1 \times 9} = \frac{25}{4 \times 9} = \frac{25}{36}$

Suggest Corrections

Similar questions

Evaluate

{lim}_{x \to 0} \frac{1 - cos 4 x}{1 - cos 5 x}

lim x \to 0 \frac{1 - cos x}{x (2^{x} - 1)}

is equal to

$lim x \to 0 \frac{1 - c o s x}{x^{2}} i s - - - - - - - - - - - - -$

lim x \to 0 \frac{1 - cos x}{x^{2}} i s e q u a l t o

lim x \to 0 \frac{1 - cos x}{x^{2}} =

Join BYJU'S Learning Program

limx→01−cos5x1−cos6x

limx→01−cos5x1−cos6x =limx→02sin25x2limx→02sin23x=2⎛⎜⎝limx→0sin5x25x2⎞⎟⎠2×254x22(limx→0sin3x3x)2×9x2 =2×1×2542×1×9=254×9=2536

${lim}_{x \to 0} \frac{1 - c o s 5 x}{1 - c o s 6 x}$