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MIT 18.01 Single Variable Calculus (OCW)
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limit as x -> pi/2 of tanx
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\[\lim_{x \rightarrow \ {\pi }/ 2 }\tan(x)={\infty}\] |dw:1348799168450:dw|
what happens when x -> pi/2
\[[\frac {sinx}{cosx} - \frac 1 {\cos(x)}] \] \[ = \frac {sinx -1}{cosx.} \ Multiply \ \top \ and \ bottom \ by \ sinx \ + \ 1:\]= (sin^2 x - / [cosx ( sinx + 1)] = - \[\frac {\cos^2 x} { [cosx ( sinx + 1)]} \frac { cosx}{ [ sinx + 1]} \ = \] which approaches\[\frac {0}{2} = 0 \]
1/0 using lopitals rule -cosx/sinx 0/1 answer is 0
as x tends to0 the equations tends to infinity
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|dw:1353170074636:dw| wow its actually undefined cos from the left its \[x \rightarrow \pi/2^- \rightarrow \infty\] \[x \rightarrow \pi/2^+\rightarrow -\infty\]
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