Compute trigonometric limit without use of de L'Hospital's rule

Question

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

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.

Answer 1

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

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

Answer 2

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

Answer 3

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

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

Answer 4

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.

Answer 5

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

Answer 6

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.

Answer 7

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