Question

# sin (T-x)15, lim

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## Let the function be , f( x )= sin( π−x ) π⋅( π−x ) We have to find the value of function at limit x→π . So we need to check the function by substituting the value at particular point ( π ), so that it should not be of the form 0 0 . If the condition is true, then we need to simplify the term to remove 0 0 form. f( x )= sin( π−π ) π⋅( π−π ) = 1 π ⋅ sin0 0 = 1 π ⋅ 0 0 = 0 0 Here, we see that the condition is not true and it is in 0 0 form. The given expression f( x )= lim x→π sin( π−x ) π⋅( π−x ) can be solved by assumption. Let the value of ( π−x ) be y . (1) According the given limits of x , x→π , then y→0 Also from equation 1, x=( π−y ) On substituting the value of new limit in terms of y , the new expression is: lim y→0 siny ( π⋅y ) (2) According to the trigonometric theorem, lim x→0 sinx x =1 (3) With the help of equations 2 and 3, we can calculate the value of limits; lim y→0 siny ( π⋅y ) = lim y→0 1 π ⋅ siny ( y ) = 1 π lim y→0 siny ( y ) On further simplification and using equation 3, we get lim y→0 siny ( π⋅y ) = 1 π ⋅1 = 1 π Thus, the value of the given expression lim x→π sin( π−x ) π⋅( π−x ) = 1 π

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