Question

lim x tends to 0 (x-sinx/tan²x)​

Answer 1

Answer:

$\lim_{x \to 0} \frac{ x-sinx}{tan^{2}x }$ = 0

Step-by-step explanation:

We need to calculate $\lim_{x \to 0} \frac{ x-sinx}{tan^{2}x }$

On putting the limit in the given question, we get $\frac{0-0}{0^{2} }$ = $\frac{0}{0}$

which is an indeterminant form.

Therefore we apply L-Hospital's Rule

=> $\lim_{n \to 0}$ $\frac{1-cosx}{2tanxsec^{2}x }$

= $\lim_{n \to 0}$ $\frac{1-cosx}{2sinxcosx }$  (As tan x = $\frac{sinx}{cosx}$ and sec x = $\frac{1}{cos x}$)

We know by standard limit results that $\lim_{x \to 0} \frac{ sinx}{x }$ = 1

Therefore multiplying and dividing the denominator by x

=> $\lim_{n \to 0}$ $\frac{1-cosx}{2sinxcosx * \frac{x}{x} }$

= $\lim_{n \to 0}$ $\frac{1-cosx}{2xcosx }$  (Using $\lim_{x \to 0} \frac{ sinx}{x }$ = 1)

On putting the limit in the given question, we get $\frac{1-1}{2*0*1 }$ = $\frac{0}{0}$

which is an indeterminant form.

Therefore we apply L-Hospital's Rule

=>  $\lim_{n \to 0}$ $\frac{-sinx}{2x(-sinx) + cos x }$

Now on putting the limit, that is x = 0, we get,

$\frac{0}{2*0*0 + 1 }$

= $\frac{0}{1}$           (As this is not an indeterminant form)

= 0

Therefore, $\lim_{x \to 0} \frac{ x-sinx}{tan^{2}x }$ = 0

Answer 2

Answer:

0.

Step-by-step explanation