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dgamma3
How would you solve the following limit: limit (x -> 0+) (x-11)/sin(x)
you can do it manually, and figure out its negative infinity. but is there any algebraic way.
\[\lim_{x \rightarrow 0+} (x-11)/\sin(x)\]
\[\lim_{x \rightarrow 0}(sinx/x)=1\]
\[You know \lim_{x \rightarrow 0+} \frac{\sin x}{x}=1 \]
so: do the same trick I told you before: Divide the top and bottom by x
and \[\lim_{x \rightarrow 0}(11/sinx)\rightarrow \infty \]
\[\lim_{x \rightarrow 0+} \frac{\frac{11-x}{x}}{\frac{\sin x}{x}} = \lim_{x \rightarrow 0+} \frac{11/x-1}{\frac{\sin x}{x}}=\]\[\frac{\infty-1}{1}=\infty\]