Question

lim x tends to 0
$\frac{ \sin( {x}^{0} ) }{x}$

​

Answer 1

$\large\underline{\sf{Solution-}}$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{sinx \degree}{x}$

We know,

$\rm :\longmapsto\:1 \degree \: = \dfrac{\pi}{180}$

So,

$\rm :\longmapsto\:x \degree \: = \dfrac{\pi \: x}{180}$

Thus,

Given limit can be rewritten as

$\rm \:  =  \:  \: \displaystyle \lim_{x \to 0} \rm \:\dfrac{sin\dfrac{\pi \: x}{180} }{x}$

$\rm \:  =  \:  \: \displaystyle \lim_{x \to 0} \rm \:\dfrac{sin\dfrac{\pi \: x}{180} }{\dfrac{\pi \: x}{180}} \times \dfrac{\pi \: }{180}$

$\rm \:  =  \:  \: \dfrac{\pi \: }{180} \: \times \: \displaystyle \lim_{x \to 0} \rm \:\dfrac{sin\dfrac{\pi \: x}{180} }{\dfrac{\pi \: x}{180}}$

$\rm \:  =  \:  \: \dfrac{\pi \: }{180} \: \times \:1$

$\red{\bigg \{ \because \: \displaystyle \lim_{x \to 0} \rm \:\dfrac{sinx}{x} = 1 \bigg \}}$

$\rm \:  =  \:  \: \dfrac{ \: \pi \: }{180}$

Hence,

$\bf :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{sinx \degree}{x} = \: \dfrac{ \: \: \pi \: \:}{180}$

Additional Information :-

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{sinx}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{tanx}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{tan^{ - 1} x}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{sin^{ - 1} x}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{log(1 + x)}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{ {e}^{x} - 1}{x} = 1$

$\rm :\longmapsto\:\displaystyle \lim_{x \to 0} \rm \: \: \dfrac{ {a}^{x} - 1}{x} = log(a)$