Question

Prove that,

$\displaystyle \lim_{x\to 0}\frac{x-log(1+x)}{1-cosx}=1$​

Answer 1

SOLUTION :-

Answer

Find the following limit:

lim_(x->0) (x-log(x+1))/(1-cos(x))

Applying l'Hôpital's rule, we get that

lim_(x->0) (x-log(x+1))/(1-cos(x)) | = | lim_(x->0) ( d/( dx)(x-log(x+1)))/( d/( dx)(1-cos(x)))

| = | lim_(x->0) (1-1/(x+1))/(sin(x))

| = | lim_(x->0) x/((x+1) sin(x))

lim_(x->0) x/(sin(x) (x+1))

By the product rule,

lim_(x->0) x/((x+1) sin(x)) = (lim_(x->0) x/(sin(x)))/(lim_(x->0) (x+1)):

1/(lim_(x->0) (x+1)) lim_(x->0) x/(sin(x))

x+1 is a polynomial and thus everywhere continuous, so

lim_(x->0) (x+1) = 1+0 = 1:

(lim_(x->0) x/(sin(x)))/(1)

Applying l'Hôpital's rule, we get that

lim_(x->0) x/(sin(x)) = lim_(x->0) (( dx)/( dx))/( d/( dx) sin(x)) = lim_(x->0) 1/(cos(x))

lim_(x->0) 1/(cos(x))

Applying the quotient rule, write lim_(x->0) 1/(cos(x)) as (lim_(x->0) 1)/(lim_(x->0) cos(x)):

1/(lim_(x->0) cos(x))

Since cosine is a continuous function,

lim_(x->0) cos(x) = cos(0) = 1:

1/1

1/1 = 1=R.H.S

Answer 2

Answer: 1

Step-by-step explanation:

Given,

$\lim_{x\to0}\frac{x-\log{(1+x)}}{1-\cos x}\\\;\\\text{Putting}\;\cos x=1-2\sin^2\frac{x}{2}\\\;\\\lim_{x\to0}\frac{x-\log(1+x)}{1-1+2\sin^2\frac{x}{2}}\\\;\\=\lim_{x\to0}\frac{x-\log(1+x)}{2\sin^2\frac{x}{2}}\\\;\\=4\lim_{x\to0}\frac{x-[x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+.........]}{2x^2\frac{\sin^2\frac{x}{2}}{\frac{x^2}{4}}}\\\;$

$=2\lim_{x\to0}\frac{\frac{x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}........}{x^2}.\lim_{x\to0}\frac{1}{\frac{\sin\frac{x}{2}}{\frac{x}{2}}.\frac{\sin\frac{x}{2}}{\frac{x}{2}}}\\\;\\=2\lim_{x\to0}\frac{\frac{1}{2}-\frac{x}{3}+\frac{x^2}{4}....}{1}.1\\\;\\=2(\frac{1}{2}-0+0.....)\\\;\\=1$

$Note:-\\\;\\1)e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}............\frac{x^n}{n!}\\\;\\2)\log(1+x)= x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}......$