Break limit

$\lim_{x\to0}\frac{\sin5x}{2x}=\frac52.\lim_{5x\to0}\frac{\sin5x}{5x}=\frac52$

$\lim_{x\rightarrow 0}\frac{sin(5x)}{2x}=\frac{5}{2}\lim_{x\rightarrow 0}\frac{sin(5x)}{5x}$

Let $y=5x$

$\therefore \frac{5}{2}\lim_{x\rightarrow0 }\frac{sin(5x)}{5x}=\frac{5}{2}\lim_{y\rightarrow0}\frac{sin(y)}{y}=\frac{5}{2}\cdot 1=\boxed{\frac{5}{2}}$

Using L'hopitals it comes out easily.

Let $f(x) = \sin 5x$ , then $\displaystyle \lim_{x \to 0} \dfrac{\sin 5x}{2x} = \lim_{\delta x \to 0} \dfrac{1}{2} \times \dfrac{\sin (5(0) + \delta x)}{0+\delta x} = f'(0) = \dfrac{5}{2}\cos 0 = \boxed{2.5}$

$lim_{x\rightarrow 0} \frac{\sin{5x}}{2x} = lim_{x\rightarrow 0}\frac{\frac{d}{dx}\sin{5x}}{2\frac{dx}{dx}}=lim_{x\rightarrow 0}\frac{5\cos{5x}}{2}=\frac{5\cos{0}}{2}=\frac{5}{2}$ Please somebody tell me how to write the limits better:'(

By l'hopitals rule

Limit Sin5x/2x=sin(3x+2x)/2x X tends to 0 = Limit Sin3xcos2x/3+cos3xsin2x/3 X tends to 0 Multiplying 3x in numerator and denominator in the first part we get = Limit Sin3xcos2x 3x / 3x 2 + cos3xsin2x/2x X tends to 0 As x tends to zero Sin3x/3x=1 sin2x/2x=1 Therefore = 1 1 3/2 +1 =5/2 =2.5