If f(9) = 9 , f′(9) = 4 , then find the value of given limit: f(x)-3x

According to the problem, we are given that f(9)=9 , f′(9)=4 and we need to find the value of the limit f(x)-3x .Let us assume L = f(x)-3x .⇒ L = f(9)-39 ⇒ L = f(9)-33-3 ⇒ L = 3-30⇒ L = 00, which is an indeterminate form.So, we can make use of the L-Hospital rule.Let us recall the L-Hospital rule. We know that if g(x)h(x) . comes out to be indeterminate form (00, ∞∞,....), then the given limit can be found as g(x)h(x) =g'xh'(x)=g"(x)h"(x) Now, let us apply the L-Hospital rule to the given limit.So, we have L = ddx(fx-3)ddx(x-3) .⇒ L = ddxf(x)-ddx3ddx(x)-ddx(3) .We know that ddx(x )=12x and dcdx = 0, where c is a constant.⇒ L = 12f(x)ddx fx-012x-0 .⇒ L = f'(x)2f(x)12x ⇒ L = f'(9)2f(9)129 .⇒ L = 4913⇒ L = 4313⇒ L = 4So, we have found the value of the given limit as 4.Therefore, The correct option for the given problem is 1.

If f(9) = 9 , f′(9) = 4 , then find the value of given limit: f(x)-3x

According to the problem, we are given that f(9)=9 , f′(9)=4 and we need to find the value of the limit f(x)-3x .Let us assume L = f(x)-3x .⇒ L = f(9)-39 ⇒ L = f(9)-33-3 ⇒ L = 3-30⇒ L = 00, which is an indeterminate form.So, we can make use of the L-Hospital rule.Let us recall the L-Hospital rule. We know that if g(x)h(x) . comes out to be indeterminate form (00, ∞∞,....), then the given limit can be found as g(x)h(x) =g'xh'(x)=g"(x)h"(x) Now, let us apply the L-Hospital rule to the given limit.So, we have L = ddx(fx-3)ddx(x-3) .⇒ L = ddxf(x)-ddx3ddx(x)-ddx(3) .We know that ddx(x )=12x and dcdx = 0, where c is a constant.⇒ L = 12f(x)ddx fx-012x-0 .⇒ L = f'(x)2f(x)12x ⇒ L = f'(9)2f(9)129 .⇒ L = 4913⇒ L = 4313⇒ L = 4So, we have found the value of the given limit as 4.Therefore, The correct option for the given problem is 1.