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# ISI MStat PSA 2019 Problem 17 | Limit of a function

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

ISI MStat 2019 PSA Problem 17

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

## Try with Hints

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 .

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

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

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

Now calculate the limit using Limit form of the Derivative.

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

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

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

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