Resolved: "Limits with variable substitution and Triglimx→0sin2(3x)x2cos⁡x=limu→0sin2(u)(u3)2=9(limu→0sin⁡uu)2"

kembdumatxf

Answered question

2022-03-23

Limits with variable substitution and Trig

lim_{x \to 0} \frac{\sin^{2} (3 x)}{x^{2} \cos x} = lim_{u \to 0} \frac{\sin^{2} (u)}{{(\frac{u}{3})}^{2}} = 9 {(lim_{u \to 0} \frac{\sin u}{u})}^{2}

Riya Erickson

Beginner2022-03-24Added 12 answers

Hint:

lim_{x \to 0} \frac{\sin^{2} (3 x)}{x^{2} \cos x} =

lim_{x \to 0} \frac{\sin (3 x)}{x} . lim_{x \to 0} \frac{\sin (3 x)}{x} . lim_{x \to 0} \frac{1}{\cos x} =

lim_{x \to 0} 3 \frac{\sin (3 x)}{3 x} . lim_{x \to 0} 3 \frac{\sin (3 x)}{3 x} . lim_{x \to 0} \frac{1}{\cos x} =

9 \underset{1}{\underset{⏟}{lim_{x \to 0} \frac{\sin (3 x)}{3 x}}} . \underset{1}{\underset{⏟}{lim_{x \to 0} \frac{\sin (3 x)}{3 x}}} . lim_{x \to 0} \frac{1}{\cos x} =

9.1 .1 . lim_{x \to 0} \frac{1}{\cos x} = 9.1 .1 .1

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