# trigonometric limit using identities

"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

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

• The second identity is not true. $\lim_{x\to0}\frac{1-\cos{(x)}}{x}=0$ – Peter Foreman Apr 10 '19 at 13:33
• @PeterForemanyou're right!!! sorry I just edited it – rorod8 Apr 10 '19 at 14:39

Hint:

Just write

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

• oh, ok so after multiplying by 1 (4/4) I then just past the lower part (1/4) to one of the other two factors, $$\frac{6+5x}{4}$$ then plugging 0 in gives $$1*1*\frac{3}{2}$$ thank you very much, very useful hint – rorod8 Apr 10 '19 at 14:42
• Yes. You can (almost) do that. But note that you cannot plug in $0$ in $\frac{4x}{\sin (4x)}$. But you can take the limit which is $1$. In the other expressions you can plug in $0$ indeed. – trancelocation Apr 10 '19 at 14:43

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

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 $$\lim_{x\to0} \frac{\frac32+\frac54x}{\left(\frac{\tan{(4x)}}{4x}\right)}=\frac32$$

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