Already have an account? Get multiple benefits of using own account!
Login in your account..!
Remember me
Don't have an account? Create your account in less than a minutes,
Forgot password? how can I recover my password now!
Enter right registered email to receive password!
As x tends to zero the value of 1/x tends to either ∞ or -∞. In this situation we will not be sure about the exact value of 1/x. As a result we will not be sure about the exact/approaching value of sin(1/x). We cant say anything about the value of sine function unless we know the angle and in this question we are not sure about the angle as at infinity it can take any value. We will be sure that the value of sin(1/x) will lie in [-1, 1] but not sure about a unique value. As in limits, it exists only when we get a unique value. Therefore we will say that the limit does not exist.
Solve for x , y (x + y - 8)/2 =( x + 2 y - 14)/3 = (3 x + y - 12 )/ 11 (Ans: x=2, y=6) Ans : x+ y - 8/2 = x + 2y - 14 /3 = 3x+ y- 12/11
what is 15,909 in roman numeral
Characteristics of Time Series Time series has the given characteristics. a) A long term trend (T) -tendency of the whole series to fall and rise. b) Seasonal variati
1/2 + 2/8 =
1. Write two m-files, one for the bisection method and another for Newton's method. 2. Using both the Bisection method and the Newton method answer the following: Include th
I need marketing management sample assignment as a guide
What is the structure of produt promotion?
help me please .76
how to describe the locus of the equation x^2+6xy+y^2+z^2=1 in cylindrical polar coordinates?
area of curve
Limit sin(1/x) when x tends to 0 is not definedCan be proved simply by multiplying and dividing by x then xsin(1/x)/x becomes 1/x as xsin(1/x)or for that matter sin(1/x)/1/x = 1 and limit reduces to 1/x which doesnt exist Also the proof can be that when x approcashes 0 from positive side 1/x tends to positive infinty and limit (right0 becomes sin(infinity) but when from left side 1/x tends to negative infinty so limit becomes -sin(infinit) which both can never b equal. so limit doesnt exist
Limit sin(1/x) when x tends to 0 is not definedCan be proved simply by multiplying and dividing by x then xsin(1/x)/x becomes 1/x as xsin(1/x)or for that matter sin(1/x)/1/x = 1 and limit reduces to 1/x which doesnt exist Also the proof can be that when x approcashes 0 from positive side 1/x tends to positive infinty and limit (right0 becomes sin(infinity) but when from left side 1/x tends to negative infinty so limit becomes -sin(infinit) which both can never b equal.
so limit doesnt exist
Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!
whatsapp: +1-415-670-9521
Phone: +1-415-670-9521
Email: [email protected]
All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd