Question

lim X approaches to zero
$\frac{x + \tan(4x) }{2x}$
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Answer 1

Answer:

$\large\boxed{\lim\limits_{x\to0}\dfrac{x+\tan(4x)}{2x}=2\dfrac{1}{2}=2.5}$

Step-by-step explanation:

$\lim\limits_{x\to0}\dfrac{x+\tan(4x)}{2x}=\lim\limits_{x\to0}\bigg(\dfrac{x}{2x}+\dfrac{\tan(4x)}{2x}\bigg)\\\\=\lim\limits_{x\to0}\dfrac{x}{2x}+\lim\limits_{x\to0}\dfrac{\tan(4x)}{2x}=\lim\limits_{x\to0}\dfrac{1}{2}+\lim\limits_{x\to0}\dfrac{\tan(4x)}{2x}\\\\=\dfrac{1}{2}+\lim\limits_{x\to0}\dfrac{\tan(4x)}{2x}=(*)$

$\lim\limits_{x\to0}\dfrac{\tan(4x)}{2x}\qquad\text{we have}\ \left(\dfrac{0}{0}\right),\text{use L'Hospital's rule}\\\\=\lim\limits_{x\to0}\dfrac{\left(\tan(4x)\right)'}{(2x)'}\\\\\bigg(\tan(4x)\bigg)'=\dfrac{1}{\cos^2(4x)}\cdot(4x)'=\dfrac{4}{\cos^2(4x)}\\\\(2x)'=2\\\\\text{substitute}\\\\=\lim\limits_{x\to0}\dfrac{\frac{4}{\cos^2(4x)}}{2}=\lim\limits_{x\to0}\dfrac{4}{2\cos^2(4x)}=\lim\limits_{x\to0}\dfrac{2}{\cos^2(4x)}=\dfrac{2}{\cos^2(4\cdot0)}\\\\=\dfrac{2}{\cos^20}=\dfrac{2}{1}=2$

$(*)=\dfrac{1}{2}+2=2\dfrac{1}{2}$