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  • If <img alt="f(a)=3, f^{\prime}(a)=-2, g(a)=-1, g^{\prime}(a)=4, \text { then } \lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}=" src="https://entrancecorner.oncodecogs.com/gif.latex?f

If f(a)=3, f^{\prime}(a)=-2, g(a)=-1, g^{\prime}(a)=4, \text { then } \lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a}=

Option: 1

-5


Option: 2

10


Option: 3

-10


Option: 4

5


Answers (1)

best_answer

\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a} We add and subtract g(a) f(a) in numerator

$$ \begin{aligned} & =\lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(a)+g(a) f(a)-g(a) f(x)}{x-a}=\lim _{x \rightarrow a} f(a)\left[\frac{g(x)-g(a)}{x-a}\right]-\lim _{x \rightarrow a} g(a)\left[\frac{f(x)-f(a)}{x-a}\right] \\ & =f_{x \rightarrow a}\left[\frac{g(x)-g(a)}{x-a}\right]-g(a) \lim _{x \rightarrow a}\left[\frac{f(x)-f(a)}{x-a}\right]=f(a) g^{\prime}(a)-g(a) f^{\prime}(a) \quad \text { [by using first principle } \\ & \text { formula] } \\ & =3.4-(-1)(-2)=12-2=10 \\ & \text { Trick: } \lim _{x \rightarrow a} \frac{g(x) f(a)-g(a) f(x)}{x-a} \end{aligned} $$ Trick:

= 3.4 – (–1)(–2)   = 12 – 2 = 10

Using L–Hospital’s rule, Limit 

\text { }=\lim _{x \rightarrow a} \frac{g^{\prime}(x) f(a)-g(a) f^{\prime}(x)}{1}

Limit =g^{\prime}(a) f(a)-g(a) f^{\prime}(a)=(4)(3)-(-1)(-2)=12-2=10 \text {. }

 

 

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