Sinar+bx20. lima,b, a + b#0
Let the function be,
f ( x ) = ( sin a x + b x ) ( a x + sin b x )
We have to find the value of the function at limit x → 0 .
So we need to check the expression by substituting the value at particular point (0), 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 a ⋅ 0 + b ⋅ 0 a ⋅ 0 + b ⋅ 0 = sin 0 0 = 0 0
Here, we see that the condition is not true and it is in 0 0 form.
So we need to reduce the function to its simplest and standard form.
f ( x ) = sin a x + b x a x + sin b x = ( sin a x a x ) ⋅ a x + b x a x + b x ⋅ ( sin b x b x ) (1)
According to the trigonometric theorem,
lim x → 0 sin x x = 1 (2)
From the theorem of limits, we know that for any two functions f and g , such that both lim x → a f ( x ) and lim x → a g ( x ) exist, then
lim x → a f ( x ) g ( x ) = lim x → a f ( x ) l i m x → a g ( x ) (3)
From equation 3, we get
lim x → 0 ( sin a x a x ) ⋅ a x + b x a x + b x ⋅ ( sin b x b x ) = lim x → 0 [ ( sin a x a x ) ⋅ a x + b x ] lim x → 0 [ a x + b x ⋅ ( sin b x b x ) ]
Thus, on expanding the equation and applying proper limits to all terms, we get
lim x → 0 [ ( sin a x a x ) ⋅ a x + b x ] lim x → 0 [ a x + b x ⋅ ( sin b x b x ) ] = lim x → 0 ( sin a x a x ) ⋅ lim x → 0 a x + lim x → 0 b x lim x → 0 a x + lim x → 0 b x ⋅ lim x → 0 ( sin b x b x )
As x → 0 , so a x → 0 and similarly b x → 0 .
Again on solving the above expression using equation 2,
lim a x → 0 ( sin a x a x ) ⋅ lim x → 0 a x + lim x → 0 b x lim x → 0 a x + lim x → 0 b x ⋅ lim b x → 0 ( sin b x b x ) = 1 ⋅ lim x → 0 a x + lim x → 0 b x lim x → 0 a x + 1 ⋅ lim x → 0 b x = lim x → 0 a x + lim x → 0 b x lim x → 0 a x + lim x → 0 b x
On taking the limit factor common from both numerator and denominator, we get
lim x → 0 ( a x + b x ) lim x → 0 ( a x + b x ) = lim x → 0 ( a x + b x a x + b x ) = lim x → 0 ( 1 ) = 1
Thus, the value of the given expression lim x → 0 ( sin a x + b x ) ( a x + sin b x ) = 1 .
Let the function be,
We have to find the value of the function at limit
So we need to check the expression by substituting the value at particular point (0), so that it should not be of the form
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
So we need to reduce the function to its simplest and standard form.
According to the trigonometric theorem,
From the theorem of limits, we know that for any two functions
From equation 3, we get
Thus, on expanding the equation and applying proper limits to all terms, we get
As
Again on solving the above expression using equation 2,
On taking the limit factor common from both numerator and denominator, we get
Thus, the value of the given expression
