Question

Can someone solve this question ~

$\[ \lim_{x\to0} \dfrac{ \sin(2x) + \sin(3x) }{2x + \sin(3x) }$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation \;\; Required\: ~}}$
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Answer 1

The limit of the given function as "x" approaches zero is 5/3

Limits of a function

Given the limit of a function expressed as:

• $\lim_{x \to 0} \frac{sin(2x)+sin(3x)}{2x+sin(3x)}$

Applying l'hospital rule

Substitute the value of x into the function to have:

$\lim_{x \to 0} \frac{sin(2x)+sin(3x)}{2x+sin(3x)} \\ =\frac{sin(2(\infty))+sin(3(0))}{2(0)+sin(3(0))} \\ =\frac{0}{0} (ind)$

Apply the l'hospital rule by differentiating the function as shown:

$\lim_{x \to 0} \frac{2cos(2x)+3cos(3x)}{2+3cos(3x)} \\=\frac{2cos(2(0))+3cos3(0))}{2(0)+3cos(3(0))} \\=\frac{2+3}{3}\\ =\frac{5}{3}$

Hence the limit of the given function as "x" approaches zero is 5/3