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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.

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babemcnuggets
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4 Answers4

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\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*}

user284331
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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.

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

Michael Rozenberg
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  • This has to do with Taylor Expansion which we did not have yet. To be precise, you look at the Expansion and use the rules for lim on the denominator and numerator? – babemcnuggets Feb 20 '18 at 04:49
  • @babemcnuggets Yes, that's what he did. A Maclaurin expansion (which is just a Taylor expansion around $x=0$) is just $\frac{f(0)}{0!} + \frac{f'(0)}{1!} + \frac{f''(0)}{2!} \cdots$, and to get $x \cos x$ and $x \sin x$, you just multiply these expansions by $x$. – Toby Mak Feb 20 '18 at 04:53
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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

lab bhattacharjee
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