Need help with limits: lim x->0 (sin^2 (x)) /(xcosx), (sin^2(x) cosx)/(1-cosx), etc

[Necron]

Need help with limits: lim x->0 (sin^2 (x)) /(xcosx), (sin^2(x) cosx)/(1-cosx), etc

lim x->0 (sin^2 (x)) /(xcosx)

lim x->0 (sin^2(x) cosx)/(1-cosx)

limx->0 (2sinx-sin2x)/(xcosx)

 

[Deleted member 4993]

lim x->0 (sin^2 (x)) /(xcosx)

lim x->0 (sin^2(x) cosx)/(1-cosx)

limx->0 (2sinx-sin2x)/(xcosx)

What are your thoughts?

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[stapel]

lim x->0 (sin^2 (x)) /(xcosx)

Play with the ratios. You know the two basic trig-ratio limits. (here) Can this ratio be rearranged to fit what you already know? For instance:

. . . . .\(\displaystyle \dfrac{\sin^2(x)}{x\, \cos(x)}\,=\, \dfrac{\sin(x)\, \sin(x)}{x\, \cos(x)}\, =\, \left(\dfrac{\sin(x)}{x}\right)\, \left(\dfrac{\sin(x)}{\cos(x)}\right)\, =\, \left(\dfrac{\sin(x)}{x}\right)\, \left(\tan(x)\right)\)

The one "term" evaluates, and the other one is a limit you know. Apply it.

lim x->0 (sin^2(x) cosx)/(1-cosx)

limx->0 (2sinx-sin2x)/(xcosx)

Apply the same sort of method to these.

If you get stuck, please reply showing your efforts so far. Thank you!