Solving $\lim_{x\rightarrow 0} \frac{\frac{1}{1+x}- \cos x}{x}$ without L'Hôpital

Question

I don't understand the following limit $$\lim_{ x \rightarrow 0 } \frac{ \frac{1}{1+x} - \cos x }{ x }$$ Can someone please help me ?

Answer 1

$$\frac{\frac{1}{1+x}-\cos x}{x}=\frac{\frac{1-\cos x-x\cos x}{1+x}}x=\frac{2\sin ^2\frac x2-x\cos x}{x(1+x)}=\frac{\sin \frac x2}{1+x}\cdot\frac{\sin \frac x2}{\frac x2}-\frac{\cos x}{1+x}$$ Now $$\lim_{x\to 0} \frac{\sin \frac x2}{1+x}\cdot \frac{\sin \frac x2}{\frac x2}=0\cdot 1=0$$ And $$\lim_{x\to 0}-\frac{\cos x}{1+x}=-1$$ So the whole limit is $-1$.

Answer 2

write your term in the form $$\frac{1-\cos(x)-x\cos(x)}{x(1+x)}$$ and multiply numerator and denominator by $1+\cos(x)$

Answer 3

HINT:

Multiply above and below by $1+x$ to get

$$\frac{1 - \cos x - x \cos x}{(1+x)x},$$

this can be equivalently expressed as the product of two limit given by

$$ \lim_{x \rightarrow 0} \frac{1}{1+x} \; \lim_{x \rightarrow 0} \frac{1 - \cos x - x \cos x}{x}. $$

The first limit is obvious. The second can be split as follows:

$$ \lim_{x \rightarrow 0 } \frac{1}{x}(1 - \cos x)- \lim_{x\rightarrow 0 }\cos x. $$

Can you continue from here?

Answer 4

Without any sophisticated formula: $$\frac{\frac1{1+x}-\cos x}{x}=\frac{\frac1{1+x}-\frac1{1+0}}{x}-\frac{\cos x-\cos 0}{x}\xrightarrow[x\to 0]{}\Bigl(\frac1{1+x}\Bigr)'_{x=0}-\cos'0=-1.$$

Answer 5

The idea is to split the numerator by subtracting and adding $1$. The given expression is thus transformed into $$\dfrac{\dfrac{1}{1+x}-1} {x}+\frac{1-\cos x} {x} $$ and this is further simplified as $$-\frac{1}{1+x}+\frac{1-\cos x} {x} $$ and note that the second term tends to $0$ so that the desired limit is $-1$.