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$\lim\limits_{x \to 0} \frac{x(1-cosx)+sinx}{x^3}$

= $\lim\limits_{x \to 0} \frac{x(1-cosx)/x+sinx/x}{x^2}$

= $\lim\limits_{x \to 0} \frac{(1-cosx)+ \lim\limits_{x \to 0} sinx/x}{x^2}$

= $\lim\limits_{x \to 0} \frac{(1-cosx)+1}{x^2}$

= $\lim\limits_{x \to 0} \frac{(sinx)}{2x}$ (by applying L'Hospital's rule)

= 1/2

Where did I went wrong here?

When applying L'Hospital's rule since starting limits comes undefined.

Edit: L'Hospital's rule is applied wrongly in above answer. Please see modified version below:

Find $\lim\limits_{x \to 0} \frac{x(1-\frac{2cosx}{3})-\frac{sinx}{3}}{x^3}$

= $\lim\limits_{x \to 0} \frac{x/x(1-\frac{2cosx}{3})-\frac{sinx}{3x}}{x^2}$

= $\lim\limits_{x \to 0} \frac{(1-\frac{2cosx}{3})- \lim\limits_{x \to 0} \frac{sinx}{3x}}{x^2}$

= $\lim\limits_{x \to 0} \frac{(1-\frac{2cosx}{3})-1/3}{x^2}$

= $\lim\limits_{x \to 0} \frac{(\frac{2sinx}{3})}{2x}$ (by applying L'Hospital's rule)

= 1/3 (correct answer: 7/18)

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    Why do you think you went wrong? Was there a different answer given? – 2017-02-17
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    When applying L'Hospital's rule since starting limits comes undefined. – 2017-02-17
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    @abhinav12369 do you know when to apply lhospitals rule? – 2017-02-17
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    yes in 0/0 in-determinant forms. I got the mistake. Thank you. – 2017-02-17
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    And yet another misdeed of the evil rule... – 2017-02-17
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    Your fundamental mistake is replacing $(\sin x)/x$ by $1$. This step is not supported by any rule of limits. – 2017-02-17

3 Answers 3

4

First, it's not (in general) correct to take "limits within limits" and replace $\sin x/x$ by $1$ while keeping the other occurences of $x$ unchanged. Second, a limit of the type $1/x^2$ is not of the type $[0/0]$, so l'Hospital doesn't apply.

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    Please see edits. In modified version I hope I didn't make same mistake. – 2017-02-17
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    You did. Same exact mistake. – 2017-02-17
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First, and I think you've already realized this, the limit in the title blows up.

$$\lim\limits_{x \to 0} \frac{x(1-\cos x)+\sin x}{x^3} = \lim\limits_{x \to 0} \frac{x(1-1+\frac{x^2}{2}+\mathrm{O}(x^4))+x-\frac{x^3}{6} + \mathrm{O}(x^5)}{x^3} = \lim\limits_{x \to 0} \frac{x+\mathrm{O}(x^3)}{x^3} = \lim_{x\to0}\frac{1}{x^2}$$

As for the latter, your problem is at $\lim\limits_{x \to 0} \frac{\frac{x}{x}\left(1-\frac{2\cos x}{3}\right)-\frac{\sin x}{3x}}{x^2} = \lim\limits_{x \to 0} \frac{\left(1-\frac{2\cos x}{3}\right)- \lim\limits_{x \to 0} \frac{\sin x}{3x}}{x^2}$; you can't split the limit like that when you have an indeterminate form. To get that, you implicitly assume an intermediate step of

$$\lim\limits_{x \to 0} \frac{\left(1-\frac{2\cos x}{3}\right)-\frac{\sin x}{3x}}{x^2} = \frac{\lim_{x\to0}\left(\left(1-\frac{2\cos x}{3}\right)- \frac{\sin x}{3x}\right)}{\lim_{x\to0}x^2}$$

the RHS of which doesn't exist. Instead what you should do is (and while I'm not one of the posters with an ancestral grudge against l'Hôpital's rule, it makes this particular problem harder than it needs to be):

$$\lim\limits_{x \to 0} \frac{x\left(1-\frac{2\cos x}{3}\right)-\frac{\sin x}{3}}{x^3} =$$ $$\lim\limits_{x \to 0} \frac{x\left(1-\frac{2}{3} + \frac{2x^2}{2\cdot3} + \mathrm{O}(x^4)\right)-\frac{x}{3} + \frac{x^3}{6\cdot3} + \mathrm{O}(x^5)}{x^3} =$$ $$\lim\limits_{x \to 0} \frac{\frac{x}{3}-\frac{x}{3} + \frac{x^3}{3} + \frac{x^3}{18} + \mathrm{O}(x^5)}{x^3} = \frac{7}{18}$$

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    Thank you, I got the mistake. Can you please elaborate more on when not to split as sometimes it work and sometimes not. – 2017-02-17
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    You can distribute across the basic arithmetic operators if the limits of both operands converge, in the case of division, the denominator to a non-zero value. You have to take it one at a time, which is why I say you assumed that impossible intermediate step. – 2017-02-17
0

Apply L'Hospital on $$\lim_{x\to0}\dfrac{x(1-\cos x)+\sin x}{x^3}$$

to find $$\lim_{x\to0}\dfrac{1-\cos x+x\sin x+\cos x}{3x^2}=?$$

For the edited version, $$F=\lim_{x\to0}\dfrac{x\left(1-\dfrac{2\cos x}3\right)+\dfrac{\sin x}3}{x^3}=\lim_{x\to0}\dfrac{x(3-2\cos x)+\sin x}{3x^3}$$ which is of the form $\dfrac00$

So, applying L'Hospital, $$F=\lim_{x\to0}\dfrac{3-3\cos x+2x\sin x}{9x^2} =\dfrac13\lim_{x\to0}\dfrac{1-\cos x}{x^2}+\dfrac29\lim_{x\to0}\dfrac{\sin x}x$$

Now applying L'Hospital $\lim_{x\to0}\dfrac{1-\cos x}{x^2}=\lim_{x\to0}\dfrac{\sin x}{2x}=?$

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    What is wrong in steps given by me? – 2017-02-17
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    @abhinav12369, You can not set $$\lim_{x\to0}\dfrac{\sin x}x=1$$ keeping $\cos x$ as it is – 2017-02-17
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    yes in 0/0 in-determinant forms. – 2017-02-17
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    please see the edited version of question. I'm getting wrong answer. – 2017-02-17
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    @abhinav12369, Please find the updated answer – 2017-02-17