How Would I solve these two limit 1. As x approaches zero Sin(x)^2/x 2. As x approaches zero Sec(x)-1/x sec(x) This is what I have done ((1/cos(x)-1))/(x)(1/cos(x))
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Hello, homeylova223! By now, you should know that more parenthese are needed . . . Is that: . Is that: .
For the both it is the latter (sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically ((secx-1))/((x sec x))
Originally Posted by homeylova223 For the both it is the latter (sin^2(x))/(x) This one I've managed to solve using a graphic calculator but I am wondering how to do it analytically ((secx-1))/((x sec x)) For Q.1, write it as . The first is a standard limit, the second can be evaluated using direct substitution. For Q.2 This is another standard limit, but if you don't know it, try multiplying top and bottom by the top's conjugate.
Hmm For Q.2 I do not understand how you can go from ((1/cos(x)-1))/(x)(1/cos(x)) to (1-cos(x)/(cos(x)/(x)/(cosx)
Originally Posted by Prove It try multiplying top and bottom by the top's conjugate. Originally Posted by homeylova223 Hmm For Q.2 I do not understand how you can go from ((1/cos(x)-1))/(x)(1/cos(x)) to (1-cos(x)/(cos(x)/(x)/(cosx) Taking Prove It's suggestion multiply by the top's conjugate edit: correction
Last edited by e^(i*pi); Jul 13th 2011 at 02:57 PM.
Originally Posted by e^(i*pi) Not quite, you forgot a term in the last expression.
Originally Posted by homeylova223 Hmm For Q.2 I do not understand how you can go from ((1/cos(x)-1))/(x)(1/cos(x)) to (1-cos(x)/(cos(x)/(x)/(cosx) Get a common denominator for
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