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This is a beautiful problem from ISI Mstat 2019 PSA problem 17 based on limit of a function . We provide sequential hints so that you can try this.

## Limit of a function

If $f(a)=2, f'(a)=1 , g(a)=-1$ and $g'(a)=2$ , then what is

$\lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x))}{(x – a)}$ ?

• 5
• 3
• – 3
• -5

### Key Concepts

Algebraic manipulation

Limit form of the Derivative

But try the problem first…

Source

ISI MStat 2019 PSA Problem 17

Introduction to real analysis Robert G. Bartle, Donald R., Sherbert.

## Try with Hints

First hint

Try to manipulate $\frac{(g(x)f(a) – g(a)f(x))}{(x – a)}$ so that you can use the Limit form of the Derivative . Let’s give a try .

Second Hint

$\frac{(g(x)f(a) – g(a)f(x))}{(x – a)}$ =

$\frac{(g(x)f(a) –g(a)f(a) +g(a)f(a) – g(a)f(x))}{(x – a)}$ =

$f(a)\frac{g(x)-g(a)}{(x-a)} – g(a)\frac{f(x)-f(a)}{(x-a)}$ .

Now calculate the limit using Limit form of the Derivative.

Final Step

So, we have $\lim\limits_{x\to a}\frac{(g(x)f(a) – g(a)f(x)}{(x – a)}$ =

$\lim\limits_{x\to a} f(a)\frac{g(x)-g(a)}{(x-a)} – \lim\limits_{x\to a} g(a)\frac{f(x)-f(a)}{(x-a)}$ =

$f(a) g'(a) – g(a)f'(a)= 2.(2)-1.(-1)=5$.