How to solve limx→0+sin⁡xsin⁡x without the use of L'Hospital's Rule

Since sin⁡xx→1 as x→0, we have thatsin⁡xsin⁡x=xsin⁡xxxsin⁡xx=sin⁡xxsin⁡xx→1as x→0

Since f(x)=x is a continuous function for x→0+, we obtain:limx→0+sin⁡{x}sin⁡{x}=limx→0+(sin⁡{x}x⋅xsin⁡{x})=limx→0+sin⁡{x}x⋅limx→0+xsin⁡{x}=1⋅1=1

Since sin⁡xx→1 as x→0, we have thatsin⁡xsin⁡x=xsin⁡xxxsin⁡xx=sin⁡xxsin⁡xx→1as x→0

Since f(x)=x is a continuous function for x→0+, we obtain:limx→0+sin⁡{x}sin⁡{x}=limx→0+(sin⁡{x}x⋅xsin⁡{x})=limx→0+sin⁡{x}x⋅limx→0+xsin⁡{x}=1⋅1=1

Get the solution to "How to solve limx→0+sin⁡xsin⁡x without the use of..."

peggyleuwpodg

2022-03-30

How to solve

lim_{x \to 0^{+}} \frac{\sqrt{\sin x}}{\sin \sqrt{x}}

without the use of L'Hospital's Rule

uqhekekocj8f

Beginner2022-03-31Added 8 answers

Since

\frac{\sin x}{x} \to 1

x \to 0

, we have that

\frac{\sqrt{\sin x}}{\sin \sqrt{x}} = \frac{\sqrt{x} \sqrt{\frac{\sin x}{x}}}{\sqrt{x} \frac{\sin \sqrt{x}}{\sqrt{x}}} = \frac{\sqrt{\frac{\sin x}{x}}}{\frac{\sin \sqrt{x}}{\sqrt{x}}} \to 1

x \to 0

LiaisyAciskriuu

Beginner2022-04-01Added 10 answers

Since

f (x) = \sqrt{x}

is a continuous function for

x \to 0^{+}

, we obtain:

lim_{x \to 0^{+}} \frac{\sqrt{\sin {x}}}{\sin {\sqrt{x}}} = lim_{x \to 0^{+}} (\sqrt{\frac{\sin {x}}{x}} \cdot \frac{\sqrt{x}}{\sin {\sqrt{x}}}) = \sqrt{lim_{x \to 0^{+}} \frac{\sin {x}}{x}} \cdot lim_{x \to 0^{+}} \frac{\sqrt{x}}{\sin {\sqrt{x}}} = \sqrt{1} \cdot 1 = 1

Do you have a similar question?

Recalculate according to your conditions!