Question

Find the limit of: lim_{x\to 0}\frac{sin2x}{sin3x}" alt="lim_{x\to 0}\frac{sin2x}{sin3x}" align="absmiddle" class="latex-formula">
Please use the hint in the problem, but also explain what the hint is. I don't know how they derived it from the original equation of: $\frac{sin2x}{sin3x}$

Answer 1

First of all I am solving the question and then I will explain the hint...

$\sf \: lim_{x\to 0} \: \frac{sin2x}{sin3x}$

• Evaluate the limits of numerator and denominator separately.

$\sf \: lim_{x\to 0} \:( sin(2x)) \\ \sf \:lim_{x\to 0} \:( sin(3x))$

• Evaluate the limit.

$\sf \: 0 \\ \sf \: 0$

• Since the expression 0/0 is an indeterminate form, try transforming the expression.

$\sf \: lim_{x\to 0} \: (\frac{sin(2x)}{sin(3x)})$

• Multiply the fraction by 2×3x/2×3x
• Now, Here we will make the use of hint.. When we evaluated the limit we got 0/0 so now we will multiply the fraction by 2×3x because we need to simplify or we can say eationalize the denominator...

$\sf \: lim_{x\to 0} \: (\frac{sin(2x) \times 2 \times 3x}{sin(3x) \times 2 \times 3x})$

• Use the commutative property to reorder the terms.

$\sf \: lim_{x\to 0} \: (\frac{ 2 \times 3x \times sin(2x)}{3 \times 2x \times sin(3x)})$

• Separate the fraction into 3 fractions.

$\sf \: lim_{x\to 0} \: ( \frac{2}{3} \times \frac{ \sin(2x) }{2x} \times \frac{3x}{ \sin(3x) } )$

• Evaluate the limit.

$\sf \: \frac{2}{3} \times 1 \times {1}^{ - 1}$

• Simplify the expression.

$\boxed{ \tt \frac{2}{3}}$