1
Question
limx→0+((xcosx)x+(cosx)1lnx+(xsinx)x) is equal to
Open in App
Solution
The correct option is D 2+1e
limx→0+(xcosx)x
limx→0+xx.cosxx
limx→0+exlnx.cosxx
1×1
1
∴limx→0+(xcosx)x=1
------------------------------------------------------
limx→0+(xsinx)x
limx→0+xx.sinxx
limx→0+exlnx.sinxx
1×1
1
∴limx→0+(xsinx)x=1
---------------------------------------------------------
limx→0+(cscx)1lnx
limx→0+(1sinx)1lnx
y=limx→0+(1sinx)1lnx
lny=limx→0+1lnx(−ln(sinx))
lny=limx→0+11x(−cosxsinx)
lny=limx→0+(−cosx)
lny=−1
y=e−1
y=1e
∴limx→0+(cscx)1lnx=1e ------------------------------------------------------------------------------------------------------------------------------------------------------------------------
limx→0+((xcosx)x+(cscx)1lnx+(xsinx)x)
=1+1e+1
=2+1e
limx→0+(xcosx)x
limx→0+xx.cosxx
limx→0+exlnx.cosxx
1×1
1
∴limx→0+(xcosx)x=1
------------------------------------------------------
limx→0+(xsinx)x
limx→0+xx.sinxx
limx→0+exlnx.sinxx
1×1
1
∴limx→0+(xsinx)x=1
---------------------------------------------------------
limx→0+(cscx)1lnx
limx→0+(1sinx)1lnx
y=limx→0+(1sinx)1lnx
lny=limx→0+1lnx(−ln(sinx))
lny=limx→0+11x(−cosxsinx)
lny=limx→0+(−cosx)
lny=−1
y=e−1
y=1e
∴limx→0+(cscx)1lnx=1e
------------------------------------------------------------------------------------
------------------------------------------------------------------------------------
limx→0+((xcosx)x+(cscx)1lnx+(xsinx)x)
=1+1e+1
=2+1e
Suggest Corrections
0
Similar questions
View More
Join BYJU'S Learning Program
Explore more
Join BYJU'S Learning Program
