Question

lim x ->0 [(1 + x² - cosx)/ x tanx]
​

Answer 1

Answer:

$\large\boxed{\sf{\dfrac{3}{2}}}$

Step-by-step explanation:

Given limit,

$\displaystyle\lim_{x\to0} \dfrac{1 + {x}^{2} - \cos(x) }{x \tan(x) }$

Solving further, we get,

$= \displaystyle\lim_{x\to0} \dfrac{1 + {x}^{2} - \cos(x) }{x \times \dfrac{ \tan(x) }{x} \times x}$

But, we know that,

• $\displaystyle\lim_{x\to0} \dfrac{ \tan(x) }{x} = 1$

Therefore, we will get,

$= \displaystyle\lim_{x\to0} \dfrac{1 + {x}^{2} - \cos(x) }{ {x}^{2} }$

Now, applying L hospital rule, i.e, differentiatiating both numerator and denominator separately, we get,

$= \displaystyle\lim_{x\to0} \dfrac{ \frac{d}{dx}(1 + {x}^{2} - \cos x) }{ \frac{d}{dx} {x}^{2} }$

But, we know that,

• $\frac{d}{dx} 1 = 0$
• $\frac{d}{dx} {x}^{2} = 2x$
• $\frac{d}{dx} \cos(x) = - \sin(x)$

Therefore, we will get,

$= \displaystyle\lim_{x\to0} \ \dfrac{2x + \sin(x) }{2x} \\ \\ = \displaystyle\lim_{x\to0} (1 + \frac{ \sin(x) }{2x} ) \\ \\ = 1 + \dfrac{1}{2} \displaystyle\lim_{x\to0} \dfrac{ \sin(x) }{x}$

But, we know that,

• $\displaystyle\lim_{x\to0} \frac{ \sin(x) }{x} = 1$

Therefore , we get,

$= 1 + \dfrac{1}{2} \times 1 \\ \\ = 1 + \dfrac{1}{2} \\ \\ = \frac{3}{2}$

Hence, required value of limit is 3/2.