0
$\begingroup$

"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.

$\endgroup$
2
  • 1
    $\begingroup$ The second identity is not true. $\lim_{x\to0}\frac{1-\cos{(x)}}{x}=0$ $\endgroup$ – Peter Foreman Apr 10 '19 at 13:33
  • $\begingroup$ @PeterForemanyou're right!!! sorry I just edited it $\endgroup$ – rorod8 Apr 10 '19 at 14:39
4
$\begingroup$

Hint:

Just write

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

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$ – rorod8 Apr 10 '19 at 14:42
  • $\begingroup$ 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. $\endgroup$ – trancelocation Apr 10 '19 at 14:43
1
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$
0
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.