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Question Number 69809 by RAKESH MANDA last updated on 28/Sep/19

lim_(x→∞)  xsin (π/x)

limxxsinπx

Commented by mr W last updated on 28/Sep/19

lim_(x→∞)  xsin (π/x)  =πlim_(x→∞)  ((sin (π/x))/(π/x))  =π lim_(t→0)  ((sin t)/t)  =π×1  =π

limxxsinπx=πlimxsinπxπx=πlimt0sintt=π×1=π

Commented by peter frank last updated on 28/Sep/19

thank you

thankyou

Answered by $@ty@m123 last updated on 28/Sep/19

Let x=(1/y)  so that as x→∞, y→0    ∴   lim_(x→∞)     xsin (π/x)   = lim_(y→0)     ((sin πy)/y)   = lim_(y→0)     ((sin πy)/(πy))×π  =1×π  =π

Letx=1ysothatasx,y0limxxsinπx=limy0sinπyy=limy0sinπyπy×π=1×π=π

Answered by malwaan last updated on 28/Sep/19

lim_(x→∞)  ((sin(𝛑/x))/(𝛑/x))×𝛑  =𝛑×lim_(t→0) ((sint)/t) = 𝛑

limxsinπxπx×π=π×limt0sintt=π

Commented by malwaan last updated on 28/Sep/19

In general   lim_(x→∞)  axsin(b/(cx))=((ab)/c)

Ingenerallimxaxsinbcx=abc

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