please help me: find limit of h(x)=sinx+cosx/sinx−2cosx as x→π/2

allos

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find the limit of
h(x)=sinx+cosx/sinx−2cosx​
as
[FONT=MathJax_Math-italic]x​
[FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math-italic]π/[/FONT][FONT=MathJax_Main]2[/FONT]​
[/FONT]
 
Last edited:
1. find the limit of
[FONT=MathJax_Math-italic]h​
[FONT=MathJax_Main]([/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main])[/FONT][FONT=MathJax_Main]=[/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]+[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Math-italic]o[/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]i[/FONT][FONT=MathJax_Math-italic]n[/FONT][FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]−[/FONT][FONT=MathJax_Main]2[/FONT][FONT=MathJax_Math-italic]c[/FONT][FONT=MathJax_Math-italic]o[/FONT][FONT=MathJax_Math-italic]s[/FONT][FONT=MathJax_Math-italic]x[/FONT]
as
[FONT=MathJax_Math-italic]x[/FONT][FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math-italic]π[/FONT][FONT=MathJax_Main]2[/FONT]​
[/FONT]
If it is \(\displaystyle x\to\pi^2\) then in the future please make it clear: x->pi^2. LOOK AT THIS
If it is not that, then what is it?
 
find the limit of
h(x)=sinx+cosx/sinx−2cosx
as x [FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math-italic]π/[/FONT][FONT=MathJax_Main]2[/FONT]
Okay, now you've changed the equation to be as follows:

. . . . .\(\displaystyle \mbox{Find the limit: }\, \)\(\displaystyle \displaystyle \lim_{x\, \rightarrow\, \frac{\pi}{2}}\,\)\(\displaystyle \bigg(\, \sin(x)\, +\, \dfrac{\cos(x)}{\sin(x)}\, -\, 2\cos(x)\, \bigg)\)

Was this what you meant? Thank you! ;)
 
find the limit of
h(x)=sinx+cosx/sinx−2cosx​
as
[FONT=MathJax_Math-italic]x​
[FONT=MathJax_Main]→[/FONT][FONT=MathJax_Math-italic]π/[/FONT][FONT=MathJax_Main]2[/FONT]​
[/FONT]

Just replace 'x' by 'π/2' in h(x) .

Where are you having trouble??
 
check my attempts

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