Question

Limx->0(sinx-x)/xsinx

Answer 1

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\bf \:\lim_{x\to0}\dfrac{sinx - x}{xsinx}$

$\bf \:\large \red{AηsωeR } ✍$

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✏️ Concept used :-

L' Hospital Rule

• So, L'Hospital's Rule tells us that if we have an indeterminate form 0/0 or ∞/∞ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.

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$\large\underline\blue{\bold{Formula \: Used :-  }}$

$\bf \:\lim_{x\to0}\dfrac{sinx}{x} = 1$

$\bf \:\dfrac{d}{dx} sinx = cosx$

$\bf \:\dfrac{d}{dx}cosx = - sinx$

$\bf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\bf \:\dfrac{d}{dx}k = 0$

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$\large\underline\purple{\bold{Solution :- }}$

$\bf \:\lim_{x\to0}\dfrac{sinx - x}{xsinx}$

$\bf \: = \lim_{x\to0}\dfrac{sinx - x}{x \times \dfrac{sinx}{x} \times x }$

$\bf \: = \lim_{x\to0}\dfrac{sinx - x}{ {x}^{2} }$

☆Using L - Hospital Rule, we get

$\bf \:\lim_{x\to0}\dfrac{cosx - 1}{2x}$

☆Using L - Hospital Rule, we get

$\bf \: = \lim_{x\to0}\dfrac{ - sinx - 0}{2}$

$\bf \: = - \dfrac{1}{2} \lim_{x\to0}sinx$

$\bf \: = - \dfrac{1}{2} \times 0 = 0$

$\large{\boxed{\boxed{\bf{Hence, \: \bf \:\lim_{x\to0}\dfrac{sinx - x}{xsinx} = 0 }}}}$