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

2 Answers

9

Explanation:

\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

Jun 28, 2018

9.

Explanation:

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.!