Find $\lim _{x \to 0} \frac {\sin x - x\ cos x} {x \sin x}$?

Question

I saw some vague similar cases here but they are not equal to this. I used wolframalpha to find out that the limes of it is 0. Also I have a solution to this, but because I dont know if my last step is okay I'll write my solution here :

Step1: We can use L'hopital since we have 0/0 from the beginning

After applying L'hopital we get $\lim _{x \to 0} \frac {x\sin x\ } {x \cos x+sinx}$

Step2: Rewritting it as $\lim _{x \to 0} \frac {\ x \ } {1+ \ (x \ cos x)/sinx}$

Step3: making use of tanx so that we have $\lim _{x \to 0} \frac {\ x \ tanx } {1+ \ x}$ since $\frac {xcosx} {sinx}$ is equal to $\frac {x}{tanx}$

Step4: This one is the step I am not really sure I can use or is correct, but because we do not have 0/0 I'd rewrite it again as $\lim _{x\to 0} \frac {\ \ tanx } {(1+ \ x)x^{-1}}$ so that it becomes 0/0 and we can use L'hopital again.

Step5: Using L'hopital we get $\lim _{x\to 0} \frac {\frac {1} {cos^{2}x} \ } {\frac{1}{-x^{2}}}$ which is $\lim _{x\to 0} \frac {-x^{2} } {cos^{2}x}$ where we could use our calculation rules so that we have $\frac {0}{-1}$ which brings us to 0.

Answer 1

\begin{align*} \lim_{x\rightarrow 0}\dfrac{x\sin x}{x\cos x+\sin x}&=\lim_{x\rightarrow 0}\dfrac{\sin x}{\cos x+\dfrac{\sin x}{x}}\\ &=\dfrac{0}{1+1}\\ &=0. \end{align*}

Answer 2

$$\frac{\sin{x}-x\cos{x}}{x\sin{x}}=\frac{x-\frac{x^3}{6}+...-x\left(1-\frac{x^2}{2}+...\right)}{x\left(x-\frac{x^3}{6}+...\right)}=\frac{\frac{1}{3}x^3+...}{x^2-\frac{x^4}{6}+...}\rightarrow0.$$

Answer 3

So, I did this a different way, and still got the answer that is given here. It might not be as rigorous as the other answers on here, but I'll give it a shot anyway.

Instead of $\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{xsinx}{xcosx + sinx}}\specialFrac{7}{z+3}$, do it's reciprocal to get rid of the x's. So, instead, do $\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{xcosx + sinx}{xsinx}}\specialFrac{7}{z+3}$. Now, this is the same as $\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{xcosx}{xsinx}}\specialFrac{7}{z+3}$ + $\def\specialFrac#1#2{\frac{sinx}{xsinx}}\specialFrac{7}{z+3}$. This now simplifies to $\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{cosx}{sinx}}\specialFrac{7}{z+3}$ + $\def\specialFrac#1#2{\frac{1}{x}}\specialFrac{7}{z+3}$. As x approaches 0, cos(x) approaches 1 and sin(x) approaches 0. So, we have $\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{cosx}{sinx}}\specialFrac{7}{z+3}$ + $\def\specialFrac#1#2{\frac{1}{x}}\specialFrac{7}{z+3}$ = $\def\specialFrac#1#2{\frac{1}{0}}\specialFrac{7}{z+3}$ + $\def\specialFrac#1#2{\frac{1}{0}}\specialFrac{7}{z+3}$ = $\def\specialFrac#1#2{\frac{2}{0}}\specialFrac{7}{z+3}$. Since this is the answer to the reciprocal ($\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{xcosx + sinx}{xsinx}}\specialFrac{7}{z+3}$), we take its reciprocal again to get the answer to the original problem ($\lim_{x \to 0}$ $\def\specialFrac#1#2{\frac{xsinx}{xcosx + sinx}}\specialFrac{7}{z+3}$). The reciprocal of $\def\specialFrac#1#2{\frac{2}{0}}\specialFrac{7}{z+3}$ is $\def\specialFrac#1#2{\frac{0}{2}}\specialFrac{7}{z+3}$ = 0 (which is what the answer should be).

I know this is not the most rigorous because I am dealing with an indeterminate form with 0 in the denominator, but it does work.

Answer 4

Hint:

$$\dfrac{\sin x-x\cos x}{x\sin x}=x\cos x\cdot\dfrac{\tan x-x}{x^3}\cdot\dfrac 1{\dfrac{\sin x}x}$$

Now use Are all limits solvable without L'Hôpital Rule or Series Expansion