Limit as $x$ goes to zero of $\sin(2x)/(2x^2+x)$ without using L'hopital!

Question

Solve $\lim_{x\to 0} \cfrac{\sin(2x)}{2x^2+x}$ with a method other than L'hopital, thanks :)

Answer 1

Near zero, by taylor's $$ \sin(2x)\sim 2x $$ So we have $$ \lim_{x\rightarrow 0}\frac{\sin(2x)}{2x^2+x}= \lim_{x\rightarrow 0}\frac{2x}{2x^2+x}=\lim_{x\rightarrow 0}\frac{2}{2x+1}=2 $$

Answer 2

Equivalents:

$\sin2x\sim_0 2x$, $2x^2+x\sim_0 x$, hence $\;\dfrac{\sin2x}{2x^2+x}\sim_0\dfrac{2x}{x}=2$.

Answer 3

First of all, notice that $2x^{2} + x = x(2x+1)$. Hence we can rewrite the given expression as $$\frac{\sin(2x)}{2x^2 + x} = \frac{\sin(2x)}{x(2x+1)} = \frac{\sin(2x)}{x}\cdot\frac{1}{2x+1}$$ Accordingly to the fundamental trigonometric limit, it results that:

\begin{align*} \lim_{x\rightarrow 0}\left[\frac{\sin(2x)}{2x^2+x}\right] & = \lim_{x\rightarrow 0} \left[\frac{\sin(2x)}{x}\cdot\frac{1}{2x+1}\right] = \lim_{x\rightarrow 0} \left[2\cdot\frac{\sin(2x)}{2x}\cdot\frac{1}{2x+1}\right]\\ & = 2\cdot\lim_{x\rightarrow 0}\frac{\sin(2x)}{2x}\cdot\lim_{x\rightarrow 0}\frac{1}{2x+1} = 2\cdot 1 \cdot 1 = 2 \end{align*}

It is worth mentioning here that we could split the limit into two limits because both of them exist.

Answer 4

All you need for a lot of problems is $\lim_{x \to 0} \frac{\sin x}{x} = 1$. This is no exception.

However, maybe you have problems with the mechanics of how to use it. There is a way to greatly simplify the use of this: there is a function called $\mathrm{sinc}$ defined by

$$ \mathrm{sinc}(x) = \begin{cases} \frac{\sin x}{x} & x \neq 0 \\ 1 & x = 0 \end{cases} $$

The fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ can now be expressed in the much simpler form that $\mathrm{sinc}(x)$ is continuous at $0$ (and everywhere else too).

Then, when taking limits as $x \to 0$, you can rewrite all occurrences of $\sin$ in terms of $\mathrm{sinc}$:

$$ \lim_{x\to 0} \frac{\sin(2x)}{2x^2+x} = \lim_{x\to 0} \frac{\mathrm{sinc}(2x) \cdot (2x)}{2x^2+x} $$

and now the right hand side is easy to simplify; we can cancel $x$ and then the result is continuous, so we can plug in $x=0$:

$$ \lim_{x\to 0} \frac{\mathrm{sinc}(2x) \cdot (2x)}{2x^2+x} = \lim_{x\to 0} \frac{\mathrm{sinc}(2x) \cdot (2)}{2x+1} = \frac{\mathrm{sinc}(0) \cdot (2)}{1} = 2 $$

Answer 5

The limit equals

$$\lim_{x\to 0} \left [\frac{\sin (2x)}{2x}\cdot\frac{1}{x+1/2}\right ] = 1\cdot \frac{1}{1/2} = 1\cdot 2 = 2.$$