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Evaluating \mathrm{ \lim _{n \rightarrow \infty} \frac{\sin \left(x+\frac{1}{n}\right)-\sin (x)}{\sin \left(x+\frac{1}{n}\right)} }

Option: 1

0


Option: 2

5


Option: 3

-1


Option: 4

1


Answers (1)

best_answer

\mathrm{\text { If } \sin (x)=0 \text { then }}

                                                   \mathrm{\frac{\sin \left(x+\frac{1}{n}\right)-\sin (x)}{\sin \left(x+\frac{1}{n}\right)}=1 \rightarrow 1}

\mathrm{\text { for } n \rightarrow \infty \text {, otherwise } \sin \left(x+\frac{1}{n}\right) \rightarrow \sin (x) \neq 0 \text { and therefore }}

                              \mathrm{\frac{\sin \left(x+\frac{1}{n}\right)-\sin (x)}{\sin \left(x+\frac{1}{n}\right)} \rightarrow \frac{\sin (x)-\sin (x)}{\sin (x)}=0 .}

So there is no need to use L'Hospital's rule in the case \mathrm{\sin (x)=0}, but doing so would give the same result:

\mathrm{\lim _{n \rightarrow \infty} \frac{\sin \left(x+\frac{1}{n}\right)-\sin (x)}{\sin \left(x+\frac{1}{n}\right)}=\lim _{h \rightarrow 0} \frac{\sin (x+h)-\sin (x)}{\sin (x+h)} \stackrel{(\mathrm{H})}{=} \lim _{h \rightarrow 0} \frac{\cos (x+h)}{\cos (x+h)}=1 .}

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Anam Khan

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