trigonometric limit using identities

Question

"find" $$\lim\limits_{x \to 0} \frac{6x+5x^2}{\tan(4x)}$$ saso what I've tried so far is splitting the $\tan(4x)$ into $\sin(4x)/\cos(4x)$ and try to get to an identity, the ones im allowed to use as identities are

$$\lim\limits_{x \to 0} \frac{\sin (x)}{x} =1$$ $$\lim\limits_{x \to 0} \frac{1-\cos (x)}{x} =0$$

please help me find it without using l'Hopital.

Answer 1

Hint:

Just write

$$\frac{6x+5x^2}{\tan{(4x)}} = \cos{(4x)}\cdot \frac{4x}{4\sin{(4x)}}\cdot(6+5x)$$

Answer 2

Well by examining the Taylor series expansion of $\tan{(x)}$ we have another 'small angle approximation' that is $$\lim_{x\to0}\frac{\tan{(x)}}{x}=1$$ So we can solve this limit by dividing through by $4x$ giving $$\lim_{x\to0} \frac{\frac32+\frac54x}{\left(\frac{\tan{(4x)}}{4x}\right)}=\frac32$$

Answer 3

Use the fact that$$\lim_{x\to0}\frac{6x+5x^2}{\tan(4x)}=\lim_{x\to0}\frac1{\frac{\sin{(4x)}}{4x}}\times\cos(4x)\times\left(\frac32+\frac{5x}4\right).$$

Answer 4

We can use the series expansion of $\tan(x)$: $$ \tan(x)=x+\frac{1}{3}x^3+\ldots $$ Ignoring higher order infinitesimals, we get $$ \lim_{x\to 0}\frac{6x+5x^2}{4x+\frac{1}{3}\times64x^3}=\frac{6x}{4x}=1.5. $$