Question

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

Answer 1

`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`

Answer 2

`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.!