Question

evaluate lim x→0 sin^2 4x / x^2

Answer 1

We have:

$\lim_{x \to 0} \dfrac{\sin^2 4x}{x^{2} }$

We have to evaluate $\lim_{x \to 0} \dfrac{\sin^2 4x}{x^{2} }$.

Solution:

$\lim_{x \to 0} \dfrac{\sin^2 4x}{x^{2} }$

Multiplying numerator and denominator by 4, we get

= $\lim_{x \to 0} 4\dfrac{\sin^2 4x}{4x^{2} }$

= $\lim_{x \to 0} 4(\dfrac{\sin 2x}{2x })^2$

= $4\lim_{x \to 0} (\dfrac{\sin 2x}{2x })^2$

Using the limit identity:

$\lim_{x \to 0} \dfrac{\sin x}{x}=1$

= $4\lim_{x \to 0} (\dfrac{\sin 2x}{2x })^2$

= $4. (1)^2$

= 4 × 1

= 4

∴ $\lim_{x \to 0} \dfrac{\sin^2 4x}{x^{2} }$ = 4

Thus, the value of $\lim_{x \to 0} \dfrac{\sin^2 4x}{x^{2} }$ = 4.

Answer 2

SOLUTION

TO EVALUATE

$\displaystyle \sf{ \lim_{x \to 0} \: \frac{ { \sin}^{2}4x }{ {x}^{2} }}$

EVALUATION

SOLVE USING L HOSPITAL RULE

$\displaystyle \sf{ \lim_{x \to 0} \: \frac{ { \sin}^{2}4x }{ {x}^{2} }} \: \: \: \: \: \: \: \: \: \bigg( \frac{0}{0} \bigg) \: form$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ 2 \times 4 \times \sin4x \cos 4x}{ 2x }} \: \: \: \: \: \: \: \:$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ 2 \sin 8x }{x }} \: \: \: \: \: \: \: \: \: \bigg( \frac{0}{0} \bigg) \: form$

$\displaystyle \sf{ = \lim_{x \to 0} \: \frac{ 2 \times 8 \times \cos 8x }{1}} \: \: \: \: \: \: \: \: \:$

$\displaystyle \sf{ = 16 \cos 0}$

$= 16 \times 1$

$= 16$

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