Question

How do you evaluate `( 5x^9-4x^2-2x+1)/(x^7-x^6+x-1)` as x approaches 1?

Answer 1

`lim_(x->1) (5x^9-4x^2-2x+1)/(x^7-x^6+x-1)=35/2`

Explanation:

`lim_(x->1) (5x^9-4x^2-2x+1)/(x^7-x^6+x-1)`

`=lim_(x->1) (color(red)(cancel(color(black)((x-1))))(5x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+x-1))/(color(red)(cancel(color(black)((x-1))))(x^6+1))`

`=lim_(x->1) (5x^8+5x^7+5x^6+5x^5+5x^4+5x^3+5x^2+x-1)/(x^6+1)`

`=(5+5+5+5+5+5+5+1-1)/(1+1)`

`=35/2`

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Alternative method

Alternatively you can use L'Hôpital's rule in the form:

If the following hold:

• `f(x)` and `g(x)` are differentiable on an open interval containing `c`, except possibly at `c`.

• `lim_(x->c) f(x) = lim_(x->c) g(x) = 0`

• `lim_(x->c) (f'(x))/(g'(x))` exists

Then:

`lim_(x->c) f(x)/g(x) = lim_(x->c) (f'(x))/(g'(x))`

`color(white)()`
For our example, let:

`f(x) = 5x^9-4x^2-2x+1`

`g(x) = x^7-x^6+x-1`

`c = 1`

Then:

• `f(x)` and `g(x)` are differentiable on an open interval containing `c=1` with:

`f'(x) = 45x^8-8x-2`

`g'(x) = 7x^6-6x^5+1`

• `lim_(x->1) f(x) = lim_(x->1) g(x) = 0`

• `lim_(x->1) (f'(x))/(g'(x)) = (45-8-2)/(7-6+1) = 35/2` exists

So:

`lim_(x->1) f(x)/g(x) = lim_(x->1) (f'(x))/(g'(x)) = 35/2`