1
Step-by-step explanation:
[tex]\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}[/tex]
Let
[tex]f(x)= \sin x[/tex]
[tex]g(x)=x[/tex]
We are going to use L'Hopital's Rule here that states
[tex]\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}=\lim_{x \to c}\dfrac{f'(x)}{g'(x)}[/tex]
We know that
[tex]f'(x) = \cos x[/tex] and [tex]g'(x)=1[/tex]
so
[tex]\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=\lim_{x \to 0}\dfrac{\cos x}{1}=1[/tex]
