sin (T-x)15, lim
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 ⋅ sin 0 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 sin y ( π ⋅ y ) (2)
According to the trigonometric theorem,
lim x → 0 sin x x = 1 (3)
With the help of equations 2 and 3, we can calculate the value of limits;
lim y → 0 sin y ( π ⋅ y ) = lim y → 0 1 π ⋅ sin y ( y ) = 1 π lim y → 0 sin y ( y )
On further simplification and using equation 3, we get
lim y → 0 sin y ( 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
Let the function be ,
We have to find the value of function at limit
So we need to check the function by substituting the value at particular point (
If the condition is true, then we need to simplify the term to remove
Here, we see that the condition is not true and it is in
The given expression
Let the value of
According the given limits of
Also from equation 1,
On substituting the value of new limit in terms of
According to the trigonometric theorem,
With the help of equations 2 and 3, we can calculate the value of limits;
On further simplification and using equation 3, we get
Thus, the value of the given expression
