How do you find the limit of #sin^2x/(1-cosx)# as #x-0#?

Question

How do you find the limit of #sin^2x/(1-cosx)# as #x->0#?

Scarlett Allison · Accepted Answer

One method is shown below.

Use algebra, trigonometry and the continuity of #cos# at #0#. #lim_(xrarr0)sin^2x/(1-cosx) = lim_(xrarr0)(1-cos^2x)/(1-cosx)# # = lim_(xrarr0)((1-cosx)(1+cosx))/(1-cosx)# # = lim_(xrarr0)(1+cosx)# # = 1+cos(0) = 1 + 1 =2#

William Abdalla · Answer

undefined To find the limit of sin^2x/(1-cosx) as x approaches 0, we can use the concept of L'Hôpital's Rule. By applying this rule, we differentiate the numerator and denominator separately with respect to x.

Differentiating sin^2x with respect to x gives us 2sinxcosx, and differentiating 1-cosx with respect to x gives us sinx.

Now, we can evaluate the limit of the differentiated expression as x approaches 0. Substituting x=0 into the differentiated expression, we get 2(0)(1) / sin(0) = 0/0.

Since we obtained an indeterminate form, we can apply L'Hôpital's Rule again. Differentiating the numerator and denominator of the differentiated expression, we get 2cos^2x - 2sin^2x in the numerator, and cosx in the denominator.

Now, we can evaluate the limit of this new expression as x approaches 0. Substituting x=0 into the new expression, we get 2(1) - 2(0) / cos(0) = 2/1 = 2.

Therefore, the limit of sin^2x/(1-cosx) as x approaches 0 is equal to 2.

How do you find the limit of #sin^2x/(1-cosx)# as #x-&gt;0#?

How do you find the limit of #sin^2x/(1-cosx)# as #x->0#?