Evaluate $\lim_{x\to 0} \dfrac {\tan (2x) - x}{3x - \sin (x)}$

Question

Evaluate $\lim_{x\to 0} \dfrac {\tan (2x) - x}{3x - \sin (x)}$

My Attempt: $$=\lim_{x\to 0} \dfrac {\tan (2x) - x}{3x - \sin (x)}$$ $$=\lim_{x\to 0} \dfrac {\dfrac {\sin (2x)}{\cos (2x)} - x}{3x - \sin (x)}$$ $$=\lim_{x\to 0} \dfrac {\sin (2x) - x.\cos (2x)}{\cos (2x)(3x-\sin (x))}$$

Answer 1

Apply L'Hospital rule. You will get the answer as $\frac{1}{2}$

Answer 2

Without Hospital, you could do it this way, divide every term in the expression by $x$ to get: $\lim_{x\to 0} \dfrac {2\frac{\tan (2x)}{2x} - 1}{3 - \frac{\sin (x)}{x}}$. Using the well known standard limits $\lim_{x\to0}\frac{\sin(x)}{x}=\frac{\tan(x)}{x}=1$ you arrive at $1/2$. Do you see it?

Answer 3

\begin{align*} \lim_{x\rightarrow 0}\frac{\tan 2x-x}{3x-\sin x}&=\lim_{x\rightarrow 0}\frac{\dfrac{2\sin 2x}{2x\cos 2x}-1}{3-\dfrac{\sin x}{x}}\\ &=\frac{2\cdot\dfrac{1}{\cos 0}-1}{3-1}\\ &=\frac{1}{2}. \end{align*}

Answer 4

Use power series (or Taylor's Theorem, if you will). We have

$$\tan(x)=x+\frac13x^3+\dots$$ $$\sin(x)=x-\frac16x^3+\dots$$

Then $\tan(2x)=2x+\frac83x^3+\dots$ so that $\tan(2x)-x=x+\frac83x^3+\dots$
On the other hand, $3x-\sin(x)=2x+\frac16x^3+\dots$ The expression can then be written as

$$\frac{x+\frac83x^3+\dots}{2x+\frac16x^3+\dots}$$

Dividing numerator and denominator by $x$ yields

$$\frac{1+\frac83x^2+\dots}{2+\frac16x^2+\dots}$$

and letting $x\to0$ we find that the limit is $\frac12$.

Answer 5

Since $(\sin x)/x\to 1$ and $\cos x\to 1$ as $x\to 0$ we have $(\tan x)/x=((\sin x)/x)(1/\cos x)\to 1$ as $x\to 0.$

So $\tan (2x)=2x(1+f(2x))$ and $\sin x=x(1+g(x))$ where $f(2x)\to 0$ and $g(x)\to 0$ as $x\to 0.$

Therefore for $x\ne 0$ we have $$\frac {\tan (2x)-x}{3x-\sin x}=\frac {x(1+f(2x))}{x(2-g(x))}=\frac {1+f(2x)}{2-g(x)}\to \frac {1}{2} \text { as } x\to 0.$$