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.

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

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

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

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

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

Outstanding Statistics Program with Applications

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