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How do you find the limit of #(x^(2)/(sinx-x))# as x approaches 0?

Answer 1

I think that the limit does not exists but you can find the lateral ones!

If you do it directly you get the form #0/0#... We can try using de l'Hospital Rule deriving top and bottom and then apply the limit. We get: #lim_(x->0)((2x)/(cos(x)-1))=0/0# again: #lim_(x->0)((2)/(-sin(x)-0))=# in this case depending on the side you chose to approach zero you get different situations. The two lateral limits are different so for #x->0# the limit does not exists: On the other hand the lateral limits give you #lim_(x->0^+)((2)/(-sin(x)-0))=2/("a negative very small number")=-oo# #lim_(x->0^-)((2)/(-sin(x)-0))=2/("a positive very small number")=+oo#

graph{x^2/(sin(x)-x) [-10, 10, -5, 5]}

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Answer 2

Emma Aldridge

To find the limit of (x^2/(sinx-x)) as x approaches 0, we can use L'Hôpital's Rule. Taking the derivative of the numerator and denominator separately, we get (2x)/(cosx-1). Evaluating this expression as x approaches 0, we find that the limit is 0.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.