Question

lim 5sinx- xcosx/ 2tanx-3x^2
(x tends to 0)​

Answer 1

lim (x → 0) (5 sinx - x cosx)/(2 tanx - 3x²) = 2

EXPLANATION:

$\displaystyle\therefore \lim_{x\to 0}\frac{5\:sinx-x\:cosx}{2\:tanx-3x^{2}}$

$\displaystyle=\lim_{x\to 0}\frac{5\frac{sinx}{x}-cosx}{2\frac{tanx}{x}-3x}$

since $\displaystyle x\to 0\implies x\neq 0$

$\displaystyle=\frac{\displaystyle 5\lim_{x\to 0}\frac{sinx}{x}-\displaystyle \lim_{x\to 0}cosx}{\displaystyle 2\lim_{x\to 0}\frac{tanx}{x}-3\displaystyle \lim_{x\to 0}x}$

$\displaystyle=\frac{5-1}{2-0}$

= 4/2

= 2

Rules:

1. $\displaystyle\lim_{x\to 0}\frac{sinx}{x}=1$

2. $\displaystyle\lim_{x\to 0}\frac{tanx}{x}=1$

3. cos0 = 1

Answer 2

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } =2$

Solution:

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } =\lim_{x \to 0} \frac{5\frac{xsinx}{x} -xcosx}{2\frac{xtanx}{x} -3x^{2} }$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } =\lim_{x \to 0} \frac{5x(\frac{sinx}{x} -cosx)}{2x(\frac{tanx}{x} -3x) }$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } = \frac{5\lim_{x \to 0}\frac{sinx}{x} -\lim_{x \to 0}cosx}{2\lim_{x \to 0}\frac{tanx}{x} -3\lim_{x \to 0}x}$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } = \frac{5(1) -cos0}{2(1) -3(0)}$

                                        $[\lim_{x \to 0} \frac{sinx}{x}=1, \lim_{x \to 0} \frac{tanx}{x}=1]$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } = \frac{5 -1}{2}$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } = \frac{4}{2}$

$\lim_{x \to 0} \frac{5sinx-xcosx}{2tanx-3x^{2} } = 2$