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I have question. I want to solve this limit. it's $\frac{0}{0}$ so we have to change it. there is two way with two different value.

$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}$

First way:
before that we know that $\lim_{x\to 0} \frac{\sin x}{x}$ or $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2}$ is equal to 1 so:

$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 - x ^ 2 (\cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{x^2 (1 - (\cos x) ^ 2)}{x^4} = \lim_{x\to 0} \frac{(\sin x) ^ 2}{x^2} = 1$

second way:
before that we know that $\sin x \sim x - \frac{x^3}{6}$ and $\cos x \sim 1 - \frac{x^2}{2}$ so:

$\lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4} = \lim_{x\to 0} \frac{(\sin x) - (x \cos x)}{x^3} * \frac{(\sin x) + (x \cos x)}{x} = \lim_{x\to 0} \frac{x - \frac{x^3}{6} - x + \frac{x^3}{2}}{x^3} * \frac{x - \frac{x^3}{6} + x - \frac{x^3}{2}}{x} = (\frac{1}{2} - \frac{1}{6}) * 2 = \frac{2}{3}$

Update:
Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$. this in another limit:

$\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$

Which way is true? Is it possible to help me?
I'm sorry for bad English.
Thanks.

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  • $\begingroup$ You evaluated a part of the limit before you get the last step, you should treat $x$ as variable until you get the step where you can substitute $x$ value and get the result. $\endgroup$ – Pentapolis Sep 12 '16 at 23:23
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Your first way is wrong, and the mistake is that you in fact argued that

$$\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1\implies \lim_{x\to0}\sin^2x=x^2$$

and this is, of course, wrong.

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  • $\begingroup$ Why? We can use Taylor expansion again. $\sin x \sim x$ so $\lim_{x\to 0} \frac{(\sin x) ^ 2}{x ^ 2} = \lim_{x\to 0} (\frac{\sin x}{x})^2 = 1$. or it's equal to $ \lim_{x\to 0} \frac{\sin x}{x} * \lim_{x\to 0} \frac{\sin x}{x} = 1 * 1 = 1$ Why it's wrong? $\endgroup$ – Amin Sep 12 '16 at 19:23
  • $\begingroup$ Because the rightmost expression above, which is what you used in your first way, is wrong. At most you could say $\;\sin^2x\sim x^2\;$ for small values of $\;x\;$, but not passing to the limit for the sine and leaving the $\;x^2\;$ ...! You could as well say that since $\;\lim_{x\to 0} x^2\cos^2x=0\;$ , then $$\;\lim_{x\to 0}\sin^2x -x^2\cos^2x \sim \lim x^2-0\;$$ and thus get the limit doesn't exist finitely... $\endgroup$ – DonAntonio Sep 12 '16 at 20:00
  • $\begingroup$ Is it possible to explain more? We have limit and we solve like this and that's work but in this limit we can't use $\lim_{x\to0}\left(\frac{\sin x}x\right)^2=1$. this in another limit: $\lim_{x\to0} \frac{1 - \cos 2x}{x^2}= \lim_{x\to0} \frac{2 (\sin x)^2}{x^2} = 2 * \lim_{x\to0} (\frac{\sin x}{x})^2 = 2 * 1 = 2$ $\endgroup$ – Amin Sep 13 '16 at 4:20
  • $\begingroup$ Your limit above is completely right and you did it alright using other well known limits, arithmetic of limits, etc. If you had used $\;\sin^2x\sim x^2\;$ for small values of $\;|x|\;$, say to estimmate something, then that'd be fine, yet what you actually did, as remarked above is $\;\lim\limits_{x\to0}\sin^2x=x^2\;$ ...you see? In the left side we have limit, in the right we don't: this is wrong! We cannot get an expression depending directly on $\;x\;$ after we have taken and solved the limit since it is $\;x\;$ the variable that is been used in the limit ! $\endgroup$ – DonAntonio Sep 13 '16 at 6:20
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    $\begingroup$ Thanks. I'm sorry . I can't vote because I don't have enough reputation. $\endgroup$ – Amin Sep 13 '16 at 7:42
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I think this problem is very common among beginners dealing with limits. The meaning of $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ is not that you can replace $\sin x$ by $x$ as if $\sin x = x$ when $x \to 0$. The meaning of equation $(1)$ is that whenever you see the expression $\lim\limits_{x \to 0}\dfrac{\sin x}{x}$ then you can replace this entire expression by $1$. There is no more meaning to equation $(1)$ and any approach to give it more meaning will only lead to confusion (as evident from current question).

The right approach to evaluation of limits is the use of theorems dealing with limits. Such theorems include laws of algebra of limits (including the exact conditions under which they are applicable), limit of composition of functions (rule of substitution), Squeeze theorem, L'Hospital's Rule and Taylor's theorem with Peano's form of remainder.


The problem here can be easily solved by factorization of numerator and denominator as follows \begin{align} L &= \lim_{x \to 0}\frac{(\sin x)^{2} - (x\cos x)^{2}}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\frac{\sin x + x \cos x}{x}\notag\\ &= \lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\cdot\lim_{x \to 0}\left(\frac{\sin x}{x} + \cos x\right)\notag\\ &= 2\lim_{x \to 0}\frac{\sin x - x\cos x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\sin x - x}{x^{3}} + \frac{1 - \cos x}{x^{2}}\notag\\ &= 2(A + B)\notag \end{align} The second limit $B$ in last step is easily calculated as $$B = \lim_{x \to 0}\frac{1 - \cos x}{x^{2}} = \lim_{x \to 0}\frac{1 - \cos^{2} x}{x^{2}(1 + \cos x)} = \frac{1}{2}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}} = \frac{1}{2}$$ The first limit $A$ can be calculated via Taylor's series (or via L'Hospital's Rule) as follows $$B = \lim_{x \to 0}\frac{\sin x - x}{x^{3}} = \lim_{x \to 0}\dfrac{x - \dfrac{x^{3}}{6} + o(x^{3}) - x}{x^{3}} = -\frac{1}{6}$$ and hence the desired limit $L$ is given by $L = 2(A + B) = 2/3$.


Someone may ask: the factorization in the numerator is obvious (based on $a^{2} - b^{2} = (a + b)(a - b)$) but why do we factor $x^{4}$ as $x^{3}\cdot x$ and perhaps not as $x^{2}\cdot x^{2}$? The answer is that one of the factors in numerator is $(\sin x + x\cos x)$ and this when divided by $x$ gives $((\sin x)/x + \cos x)$ which has a well defined limit $2$.

The next question is: why not pair the factor $x$ in denominator with factor $(\sin x - x\cos x)$ in numerator? After all this also leads to $((\sin x)/x - \cos x)$ which has a well defined limit $0$. This gives rise to an important fact about the laws of algebra of limits especially in case of product.

The usual statement of product law of limits says that if both $\lim_{x \to a}f(x)$ and $\lim_{x \to a}g(x)$ exist then $\lim_{x \to a}f(x)g(x)$ also exists and $$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x)\tag{2}$$ Thus the rule requires us to first verify that limits of both the factors $f(x)$ and $g(x)$ exist. A more convenient improvement on the above rule is the following:

If $\lim_{x \to a}f(x)$ exists and is non-zero then the following equation $$\lim_{x \to a}f(x)g(x) = \lim_{x \to a}f(x)\cdot\lim_{x \to a}g(x)\tag{3}$$ holds irrespective of the fact whether $\lim_{x \to a}g(x)$ exists or not. The meaning of equation $(3)$ is to be interpreted in this manner: if $\lim_{x \to a}g(x)$ exists then $\lim_{x \to a}f(x)g(x)$ also exists and its value is given by equation $(3)$. If $\lim_{x \to a}g(x)$ does not exist then the behavior of $\lim_{x \to a}f(x)g(x)$ is same as that of $\lim_{x \to a}g(x)$.

The above formulation of the product rule is more helpful in practice because it requires you to check the existence of limit of only one of the factors and if this limit is non-zero you can safely use the product rule. Hence when using the product rule it always makes sense to have one factor with non-zero limit especially when you are not sure about the limit of other factor. Note however that if the limit of a factor is $0$ then you must investigate/analyze the limit of other factor before reaching any conclusion about the limit of the product.

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  • $\begingroup$ Thanks for answer. I understand that we need to see $\lim_{x \to 0}\frac{\sin x}{x} = 1$ but why we can't make it? for example the answer of this limit is: $\lim_{x \to 0}\frac{\sin 2x}{\sin x} = \lim_{x \to 0}\frac{2x \frac{\sin 2x}{2x}}{x \frac{sin x}{x}} = \lim_{x \to 0}\frac{2x}{x} = \lim_{x \to 0} 2 = 2$ but in my question I can't make it. Why? If we can't make it all the time the how can we solve $\lim_{x \to 0}\frac{\sin 2x}{sin x}$? $\endgroup$ – Amin Sep 13 '16 at 10:30
  • $\begingroup$ @Amin: we solve it like $$\lim_{x \to 0}\frac{\sin 2x}{\sin x} = \lim_{x \to 0}\frac{\sin 2x}{2x}\cdot\frac{2x}{x}\cdot\frac{x}{\sin x} = 1\cdot 2 \cdot 1 = 2$$ $\endgroup$ – Paramanand Singh Sep 13 '16 at 10:33
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The correct way is the second one. The limit is indeed $\frac{2}{3}$.

Since$x$ is approaching to zero, you're lead to use Taylor expansion which is a cool tool to evaluate some limits.

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  • $\begingroup$ Yes. true answer is second way but why first way is not true. $\endgroup$ – Amin Sep 12 '16 at 19:25
  • $\begingroup$ @Amin Your first try is wrong because you did a wrong maths. $\endgroup$ – Von Neumann Sep 12 '16 at 19:39
  • $\begingroup$ @FourierTransform I honestly think your answer to the OP's comment doesn't really help... $\endgroup$ – DonAntonio Sep 12 '16 at 20:07
  • $\begingroup$ @DonAntonio Why not? Excluding the fact that his first question was "Which way is true?" , which needs a boolean answer, the "help" is within what I've said. I mean if the limit were been, let's say, to infinity, Taylor Series couldn't be used. Or maybe it was supposed to give him also a reason why the first one was wrong? $\endgroup$ – Von Neumann Sep 12 '16 at 20:12
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    $\begingroup$ @DonAntonio I'll keep that in mind for the next time. In the meantime I'll +1 you because you helped him more than me! $\endgroup$ – Von Neumann Sep 12 '16 at 20:22
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Note that we have

$$\begin{align} \frac{\sin^2(x)-x^2\cos^2(x)}{x^4}&=\frac{\left(\frac{\sin(x)}{x}\right)^2-1}{x^2}+\frac{\sin^2(x)}{x^2}\\\\ &=\color{blue}{\underbrace{\left(\frac{\sin(x)}{x}+1\right)}_{\to 2\,\,\text{as}\,\,x\to 0}}\left(\frac{\sin(x)-x}{x^3}\right)+\color{red}{\underbrace{\frac{\sin^2(x)}{x^2}}_{\to 1\,\,\text{as}\,\,x\to 0}}\\\\ \end{align}$$

Evaluating the limit of interest boils down to evaluating the limit of $\frac{\sin(x)-x}{x^3}$.

Given that $\sin(x)-x=-\frac16 x^3+O(x^5)$, the limit is easily seen to be $-\frac16$.

Putting it all together, we find the limit of interest is

$$\lim_{x\to 0}\left(\frac{\sin^2(x)-x^2\cos^2(x)}{x^4}\right)=\frac23$$

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  • $\begingroup$ Please let me know how I can improve my answer. I really want to give you the best answer I can. -Mark $\endgroup$ – Mark Viola Oct 5 '16 at 23:39
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If you are familiar with the Maclaurin series for the basic trig functions, you can solve this limit and others like it very quickly or even in your head. It also avoids differentiation mistakes from L'Hospital's.

In your case, second order expansion is enough. Taking $$ \sin(x)\sim x-\frac{x^3}{6}\\ x\cos(x)\sim x(1-\frac{x^2}{2})=x-\frac{x^3}{2} $$ Your limit becomes $$ \lim_{x\to 0} \frac{(\sin x)^2 - (x \cos x) ^ 2}{x^4}= \lim_{x\to 0} \frac{(x- \frac{x^3}{6})^2- (x-\frac{x^3}{2}) ^ 2}{x^4} $$ And now you can notice that lower order terms terms dominate, and fortunately the $x^2$ cancel (or it would diverge), and pluck out the coefficients on $x^4$, which is $2/3$. If this is uncomfortable, you can always expand and check.

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Your first way is wrong and, unfortunately, a common mistake. With the same argument you would conclude that $$ \lim_{x\to0}\frac{x-\sin x}{x^3}=0 $$ which is a big error (see below).

You can change $\sin x$ into $x$ when it's a factor, not a summand. For instance, if you have $$ \lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x} $$ you can as well compute $$ \lim_{x\to0}\frac{\sqrt{1-x}-1}{x}= \lim_{x\to0}\frac{1-x-1}{x(\sqrt{1-x}+1)}=-\frac{1}{2} $$ because then $$ \lim_{x\to0}\frac{\sqrt{1-x}-1}{\sin x}= \lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\frac{x}{\sin x}= \lim_{x\to0}\frac{\sqrt{1-x}-1}{x}\lim_{x\to0}\frac{x}{\sin x}=-\frac{1}{2}\cdot 1 $$

For the second way you have used Taylor expansions and, indeed, the first limit above can be computed correctly with them: $$ \lim_{x\to0}\frac{x-\sin x}{x^3}= \lim_{x\to0}\frac{x-\left(x-\dfrac{x^3}{6}+o(x^3)\right)}{x^3}=\frac{1}{6} $$ Do you see? The quotient $\dfrac{\sin x}{x}$ is “like $1$”, but the difference $x-\sin x$ is “like $\dfrac{x^3}{6}$”.

You can tackle the computation of the limit by observing that \begin{align} (\sin x)^2-(x\cos x)^2 &=(\sin x-x\cos x)(\sin x+x\cos x) \\ &=\left(x-\frac{x^3}{6}+o(x^3)-x\left(1-\frac{x^2}{2}+o(x^2)\right)\right) (x+o(x)+x(1+o(1)) \\ &=\left(\frac{x^3}{3}+o(x^3)\right)(2x+o(x)) \\ &=\frac{2}{3}x^4+o(x^4) \end{align}

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The second factor in $$\frac{\sin x-x\cos x}{x^3}\frac{\sin x+x\cos x}{x}$$ clearly tends to $2$.

Then by L'Hospital,

$$\frac{\cos x-\cos x+x\sin x}{3x^2}$$ tends to $1/3$.

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