Please Help :) Lim (Ln(1+9X)+cos(4X)-1)/X X=0 ???

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Answer 1 · 2018-06-28T17:23:36

#9#

#\lim_{x\to 0}\frac{\ln(1+9x)+\cos(4x)-1}{x}#

Applying L'Hospital's rule for #0/0# form as follows

#\lim_{x\to 0}\frac{\frac{d}{dx}(\ln(1+9x)+\cos(4x)-1)}{\frac{d}{dx}x}#

#=\lim_{x\to 0}\frac{\frac{1}{1+9x}\cdot 9-4\sin(4x)}{1}#

#=\lim_{x\to 0}(\frac{9}{1+9x}-4\sin(4x))#

#=\frac{9}{1+0}-4(0)#

#=9#

Answer 2 · 2018-06-28T17:40:59

# 9#.

We will use the following Standard Limits :

#(1) : lim_(t to 0)(1+t)^(1/t)=e#.

#(2) : lim_(h to 0)sinh/h=1#.

#"Now, the Reqd. Lim."=lim_(x to 0){ln(1+9x)+cos4x-1}/x#,

#=lim1/xln(1+9x)-(1-cos4x)/x#,

#=lim(9*1/(9x))ln(1+9x)-(2sin^2 2x)/x#,

#=lim[9*ln(1+9x)^(1/(9x))-2*(2sinxcosx)^2/x]#,

#=lim[9*ln(1+9x)^(1/(9x))-8*sinx/x*sinxcos^2x]#.

Since, #ln# function is continuous, we have,

#lim_(x to 0)ln(1+9x)^(1/(9x))=ln{lim_(x to 0)(1+9x)^(1/(9x))}=lne#,

#=1#.

Hence, #"the reqd. lim"=9*1-8*1*sin0*cos^2 0#,

#=9-0#,

#=9#.

Enjoy Maths.!