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Question: If \(f(a) = 3,f^{'}(a) = - 2,g(a) = - 1,g^{'}(a) = 4,\) then \(\lim_{x \rightarrow a}\frac{g(x)f(a)...

If f(a)=3,f′(a)=−2,g(a)=−1,g′(a)=4,f(a) = 3,f^{'}(a) = - 2,g(a) = - 1,g^{'}(a) = 4, then

lim⁡x→ag(x)f(a)−g(a)f(x)x−a\lim_{x \rightarrow a}\frac{g(x)f(a) - g(a)f(x)}{x - a}=

A

– 5

B

10

C

– 10

D

5

Answer

10

Explanation

Solution

lim⁡x→ag(x)f(a)−g(a)f(x)x−a\lim_{x \rightarrow a}\frac{g(x)f(a) - g(a)f(x)}{x - a}. We add and subtract g(a)f(a)g(a)f(a) in numerator

= lim⁡x→ag(x)f(a)−g(a)f(a)+g(a)f(a)−g(a)f(x)x−a\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→af(a)[g(x)−g(a)x−a]−lim⁡x→ag(a)[f(x)−f(a)x−a]\lim_{x \rightarrow a}f(a)\left\lbrack \frac{g(x) - g(a)}{x - a} \right\rbrack - \lim_{x \rightarrow a}g(a)\left\lbrack \frac{f(x) - f(a)}{x - a} \right\rbrack

= f(a)lim⁡x→a[g(x)−g(a)x−a]−g(a)lim⁡x→a[f(x)−f(a)x−a]f(a)\lim_{x \rightarrow a}\left\lbrack \frac{g(x) - g(a)}{x - a} \right\rbrack - g(a)\lim_{x \rightarrow a}\left\lbrack \frac{f(x) - f(a)}{x - a} \right\rbrack

= f(a)g′(a)−g(a)f′(a)f(a)g'(a) - g(a)f'(a) [by using first principle formula]

= 3.4 – (–1)(–2) = 12 – 2 = 10

Trick : lim⁡x→ag(x)f(a)−g(a)f(x)x−a\underset{\mathbf{x}\mathbf{\rightarrow}\mathbf{a}}{\mathbf{\lim}}\frac{\mathbf{g(x)f(a)}\mathbf{-}\mathbf{g(a)f(x)}}{\mathbf{x}\mathbf{-}\mathbf{a}}

Using L–Hospital’s rule, Limit = lim⁡x→ag′(x)f(a)−g(a)f′(x)1\lim_{x \rightarrow a}\frac{g'(x)f(a) - g(a)f'(x)}{1};

Limit =g′(a)f(a)−g(a)f′(a)g'(a)f(a) - g(a)f'(a) = (4)(3)−(−1)(−2)(4)(3) - ( - 1)( - 2) = 12 – 2 = 10.