Question

l$\lim_{n \to \0} \frac{sin x}{x}$

Answer 1

1

Step-by-step explanation:

$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}$

Let

$f(x)= \sin x$

$g(x)=x$

We are going to use L'Hopital's Rule here that states

$\displaystyle \lim_{x \to c}\dfrac{f(x)}{g(x)}=\lim_{x \to c}\dfrac{f'(x)}{g'(x)}$

We know that

$f'(x) = \cos x$ and $g'(x)=1$

so

$\displaystyle \lim_{x \to 0}\dfrac{\sin x}{x}=\lim_{x \to 0}\dfrac{\cos x}{1}=1$