Question
Determine $\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}$Substituting $x=0$ gives$$\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}=\frac{\sin 0-0}{0}=\frac{0}{0}$$Applying L'Hôpital's rule gives$$\lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}=\lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{2 x}\right\}$$Substituting $x=0$ gives$$\frac{\cos 0-1}{0}=\frac{1-1}{0}=\frac{0}{0} \text { again }$$Applying L'Hôpital's rule again gives$$\lim _{x \rightarrow 0}\left\{\frac{\cos x-1}{2 x}\right\}=\lim _{x \rightarrow 0}\left\{\frac{-\sin x}{2}\right\}=0$$
Step 1
Step 1: Substitute $x=0$ into the limit expression: $$ \lim _{x \rightarrow 0}\left\{\frac{\sin x-x}{x^{2}}\right\}=\frac{\sin 0-0}{0}=\frac{0}{0} $$ This is an indeterminate form, so we can apply L'Hôpital's rule. Show more…
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