8
$\begingroup$

$$ \lim_{x\to 0} \frac{(x+c)\sin(x^2)}{1-\cos(x)}, c \in \mathbb{R^+} $$

Using de L'Hospital's rule twice it is possible to show that this limit equals $2c$. However, without the use of de L'Hospital's rule I'm lost with the trigonometric identities.

I can begin by showing $$ \lim\frac{x\sin (x^2)(1+\cos(x))}{\sin^2x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x}=\lim\frac{\sin (x^2)(1+\cos(x))}{\sin x}+\frac{c\sin x^2(1+\cos x)}{\sin^2x}, $$

and here I'm getting stuck. I will appreciate any help.

$\endgroup$
3
  • 1
    $\begingroup$ You were almost there, just notice that lim(sin(xˆ2)/sinˆ2(x))=1, and the second part gives 2c, the first part gives 0. $\endgroup$
    – user
    Jul 28, 2016 at 14:24
  • $\begingroup$ A notation question, relevant to at least one answer: do you consider $\mathbb{R}^+$ as the set of positive real numbers, or as the set of non-negative ones? $\endgroup$
    – Clement C.
    Jul 28, 2016 at 15:04
  • $\begingroup$ Strictly positive numbers $\endgroup$
    – J.Doe
    Jul 29, 2016 at 2:23

7 Answers 7

12
$\begingroup$

We can solve in two steps fistly multipling both sides to ${ x }^{ 2 }$ and then to $1+\cos { \left( x \right) } $ and consider the fact $\lim _{ x\rightarrow 0 }{ \frac { \sin (x^{ 2 }) }{ x^{ 2 } } =1 } $ we will get

$$\lim _{ x\to 0 } \frac { (x+c)\sin (x^{ 2 }) }{ 1-\cos (x) } =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \left( 1-\cos (x) \right) } \frac { \sin (x^{ 2 }) }{ { x }^{ 2 } } =\\ =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \left( 1-\cos (x) \right) } \frac { \left( 1+\cos (x) \right) }{ \left( 1+\cos (x) \right) } =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ 1-\cos ^{ 2 }{ \left( x \right) } } \left( 1+\cos { \left( x \right) } \right) \\ =\lim _{ x\to 0 } \frac { (x+c){ x }^{ 2 } }{ \sin ^{ 2 }{ \left( x \right) } } \left( 1+\cos { \left( x \right) } \right) =\lim _{ x\to 0 } \left( x+c \right) \left( 1+\cos { \left( x \right) } \right) =2c$$

$\endgroup$
2
  • $\begingroup$ How did you progress from the third to last to second to last equality? $\endgroup$
    – zz20s
    Jul 28, 2016 at 23:23
  • $\begingroup$ @zz20s,because of $\lim _{ x\to 0 } \frac { { x }^{ 2 } }{ \sin ^{ 2 }{ \left( x \right) } } =1$ $\endgroup$
    – haqnatural
    Jul 28, 2016 at 23:31
10
$\begingroup$

Using Taylor series make life "easy". Start with $$\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$$ $$\sin(x^2)=x^2-\frac{x^6}{6}+O\left(x^8\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ So,$$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=\frac{(x+c)\left(x^2-\frac{x^6}{6}+O\left(x^8\right) \right)}{\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)}$$ Now, long division to get $$\frac{(x+c)\sin(x^2)}{1-\cos(x)}=2 c+2 x+\frac{c x^2}{6}+\frac{x^3}{6}+O\left(x^4\right)$$ which shows the limit and how it is approached.

$\endgroup$
3
  • 1
    $\begingroup$ Maybe relevant to point out: here, you indeed obtain gow [the limit] is approached," via the lower order terms. If you only need the limit, then $\sin x = x + o(x)$ and $\cos x = 1-\frac{x^2}{2} + o(x^2)$ is all you need. $\endgroup$
    – Clement C.
    Jul 28, 2016 at 14:05
  • 1
    $\begingroup$ Once more, I totally agree with you ! Cheers :-) $\endgroup$ Jul 28, 2016 at 14:17
  • $\begingroup$ Pretty sure you don't want big O notation here. $\endgroup$ Jul 29, 2016 at 8:22
7
$\begingroup$

With equivalents, it's straightforward: $$\sin u\sim_0 u,\quad 1-\cos x\sim_0 \frac12x^2,\enspace\text{hence}\quad \frac{(x+c)\sin x^2}{1-\cos x}\sim_0\frac{cx^2}{\frac12x^2}=2c.$$

$\endgroup$
4
$\begingroup$

Your first step is very good: by multiplying numerator and denominator by $1+\cos x$, the limit becomes $$ \lim_{x\to0}(x+c)(1+\cos x)\frac{\sin(x^2)}{\sin^2 x}= \lim_{x\to0}(x+c)(1+\cos x)\frac{\sin(x^2)}{x^2}\frac{x^2}{\sin^2 x} $$ and it should be easy to finish.

$\endgroup$
3
$\begingroup$

$\dfrac{\sin(x^2)}{\sin x} = \dfrac{\sin(x^2)}{x^2} \dfrac{x^2}{x} \dfrac{x}{\sin x}$ tends to 0.

Similarly : $c\dfrac{\sin(x^2)}{\sin^2 x} = c\dfrac{\sin(x^2)}{x^2} \left( \dfrac{x}{\sin x}\right)^2$ tends to $c$.

Since $1+\cos x$ tends to 2, the result follows.

$\endgroup$
2
$\begingroup$

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

$\sin x^2=x^2G(x)$ where $\lim_{x\to 0}G(x)=1.$

$\frac {(x+c)\sin x^2}{1-\cos x}=\frac {(x+c)x^2 G(x)}{2(\frac {x}{2})^2F(x)}=(2x+2c)\frac {F(x)}{G(x)}.$

The rest is obvious.

$\endgroup$
2
$\begingroup$

Easy to see $\lim_{x\to 0}\frac{ \sin(x^2)}{sin^2 x} = 1$

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

$\endgroup$
4
  • 1
    $\begingroup$ The first line is true, but the logical connection afterwards (and hence the proof) is wrong. $\lim x=\lim x^3$, yet $\lim \frac x x \neq \lim \frac {x^3}x $ at 0. The reason your equalities hold is not the one you gave, but it's because you are implicitly using the fact the two functions are equivalent (not only that they have the same limit). $\endgroup$
    – Clement C.
    Jul 28, 2016 at 13:58
  • $\begingroup$ (Hence my downvote, incidentally: the argument is flawed.) $\endgroup$
    – Clement C.
    Jul 28, 2016 at 14:25
  • 1
    $\begingroup$ It would be better to write that $\lim_{x\to0}\frac{\sin(x^2)}{\sin^2(x)}=1$; this lets the argument progress correctly. The problem with using the statement that you do is that, since both of those limits are zero, by replacing one with the other you implicitly divide by zero; this argument would work fine if, for instance, $\lim_{x\to0}\sin(x^2)$ weren't zero. $\endgroup$ Jul 28, 2016 at 22:17
  • $\begingroup$ After the edit, indeed it's correct. $\endgroup$
    – Clement C.
    Jul 29, 2016 at 1:49

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .