Solving the following limit without L'Hospital's rule: $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x} $

Question

I have been trying to solve the following limit $$\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}.$$

I came across the right answer as shown by the steps below, but I would to check if the steps are correct or if someone has a more straightforward solution.

So applying the sum formula for sine and doing simple algebra we have: $$\lim_{x\to 0} \frac{\cos{2} \,(\sin{x^2}-\sin{x})}{x} - \frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$

The first limit is easy to evaluate and is equal to $-\cos{2}$. However, the second limit is harder, as it follows: $$\lim_{x\to 0}\frac{\sin{2} (\cos{x^2}-\cos{x})}{x} .$$

I came across a solution by using the following sum-to-product identity: $$\cos{A}-\cos{B}=-2\sin{\Big(\frac{A+B}{2}\Big)} \sin{\Big(\frac{A-B}{2}\Big)}$$

Setting $A=x^2$ and $B=x$, we have that $$\cos{x^2}-\cos{x}=-2\sin{\Big(\frac{x^2+x}{2}\Big)} \sin{\Big(\frac{x^2-x}{2}\Big)}$$

This is my only point of concern whether I applied the identity correctly. The rest of it flows more easily:

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

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

The first limit can be solved as it follows: $$\lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)}}{x} = \lim_{x\to 0} \frac{\sin{\Big(\frac{x^2+x}{2}\Big)} \Big(\frac{x^2+x}{2}\Big)}{x \Big(\frac{x^2+x}{2}\Big)} = 1 \cdot \lim_{x \to 0} \frac{x^2 + x}{2x} = \frac{1}{2} $$

The second limit is equal to zero $$\lim_{x\to 0} \sin{\Big(\frac{x^2-x}{2}}\Big)=0$$

Answer 1

You can just sum-to-product it from the beginning and use $\lim_{x\rightarrow 0}\sin(x)/x = 1$: \begin{align} \lim_{x\rightarrow 0}\frac{\sin(x^2+2)-\sin(x+2)}{x} &= \lim_{x\rightarrow 0}\frac{2\cos[2+x(x+1)/2]\sin[x(x-1)/2]}{x} \\ & = \cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x/2} \\ &= -\cos(2)\lim_{x\rightarrow 0}\frac{\sin[x(x-1)/2]}{x(x-1)/2} \\ &= -\cos(2) \end{align}

Answer 2

Hint:

You can rewrite this fraction as $$\frac{\sin(x^2+2)-\sin(x+2)}{x}=\frac{\sin(x^2+2)-\sin(2)}{x}-\frac{\sin(x+2)-\sin(2)}{x},$$ which is the difference of two rates of variation. Can take it from there?

Answer 3

The second limit is not really hard: leaving out the $\sin2$ factor, we have $$ \lim_{x\to0}\frac{\cos(x^2)-\cos x}{x}= \lim_{x\to0}\frac{\cos(x^2)-1+1-\cos x}{x} $$ It's easy to show that $$ \lim_{x\to0}\frac{1-\cos x}{x}=0 $$ so we remain with $$ \lim_{x\to0}\frac{\cos(x^2)-1}{x}=\lim_{x\to0}\frac{\cos(x^2)-1}{x^2}x $$ which is zero for the same reason.

Answer 4

First notice that

$$\lim_{x\to 0} \frac{\cos x - 1}{x} = \lim_{x\to 0} \frac{2\sin^2\frac{x}2}{x} = \lim_{x\to 0} \left(\frac{\sin^2\frac{x}2}{\frac{x}2}\right)\cdot\sin\frac{x}2 = 1 \cdot 0 = 0$$ so

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

Finally we get

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

Answer 5

On applying Taylor series, and binomial expansion and taking linear degree terms in numerator to evaluate the limit,(all constant terms are powers of 2, they cancel out as they are same for both terms in numerator)

We get the limit as $\lim_{x\to 0} \frac{\sin(x^2+2)-\sin(x+2)}{x}. = -1 + \frac{{3} \choose {1} * 2^2}{3!} - \frac{{5} \choose {1} *2^4}{5!} \ldots =-(1-\frac{2^2}{2!}+\frac{2^4}{4!} \ldots)=-\cos(2)$

Answer 6

The second limit can be evaluated as follows:

$\quad \quad \quad \lim_{x\to0} \frac{sin2(cosx^2 - cosx)}{x}$

$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2 - cosx}{x})$

$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{cosx^2}{x} - \frac{cosx}{x})$

$\quad \quad \quad \lim_{x\to0} sin2 \cdot (\frac{1}{x} \cdot (cosx^2 - cosx))$

$\quad \quad \quad sin2 \cdot \lim_{x\to0} \frac{1}{x} \cdot \lim_{x\to0} (cosx^2-cosx)$

By the multiplication rule for limits

$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot (cosx^2-cosx))$

Substituting $x = 0$ for the latter expression

$\quad \quad \quad sin2 \cdot \lim_{x\to0} (\frac{1}{x} \cdot 0)$

$\quad \quad \quad sin2 \cdot \lim_{x\to0} (0)$

The second part is the limit of a constant, which is just the constant, therefore

$\quad \quad \quad sin2 \cdot 0$

Therefore, the entire second limit evaluates to $0$, and the answer to your question is the first limit; $-cos(2)$.