Question

lim x-0 tanx - sinx / 1-cosx

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Answer 1

$\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{tan\: x - sin\:x}{1 - cos\:x}$

$\sf put \: x = 0 \: in \: \dfrac{tan\: x - sin\:x}{1 - cos\:x}$

$\implies \sf \: \dfrac{tan\: 0\degree - sin\:0\degree}{1 - cos\:0 \degree}$

We know that :-

$\underline{\boxed{\bf \tan(0) = 0 }}$

$\underline{\boxed{\bf \sin(0) = 0 }}$

$\underline{\boxed{\bf \cos(0) = 1 }}$

$\implies \sf \: \dfrac{ 0 - 0}{1 - 1}$

$\implies \sf \: \dfrac{ 0}{0}$

Therefore we get 0/0 form. Now we will simply the given expression.

We know that :-

$\underline{\boxed{\bf \tan(t) = \frac{ \sin(t) }{ \cos(t) } }}$

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{ \dfrac{sin \: x}{cos \: x} - sin\:x}{1 - cos\:x}$

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{ \dfrac{sin \: x}{cos \: x} - \dfrac{sin\:x}{1} }{1 - cos\:x}$

LCM of cos x and 1 = cos x

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: { \dfrac{sin \: x -sin\:x \:cos\:x }{cos \: x(1 - cos\:x)} }$

Taking out sin x as common :-

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: { \dfrac{sin \: x (1- \:cos\:x )}{cos \: x(1 - cos\:x)} }$

Now cancel out (1-cos x) from numerator and denominator.

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: { \dfrac{sin \: x \: \: \cancel{(1- \:cos\:x )}}{cos \: x \: \: \cancel{(1- \:cos\:x )}} }$

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: { \dfrac{sin \: x }{cos \: x } }$

Now put x = 0

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: { \dfrac{sin \: 0\degree }{cos \: 0 \degree} }$

We know :-

$\underline{\boxed{\bf \sin(0) = 0 }}$

$\underline{\boxed{\bf \cos(0) = 1 }}$

$\implies\displaystyle \sf \lim \limits_{x \to \: 0} \: \frac{0}{1} = 0$

Answer :

$\displaystyle \sf \lim \limits_{x \to \: 0} \: \dfrac{tan\: x - sin\:x}{1 - cos\:x} = 0$