Question

If `f(x)=(4x^2-7x-2)/(x-2),x!=2`,find `aδ>0` such that `|f(x)-9|<1/100` for `0<|x-2|<δ`. Hence show that `lim_(xto2)f(x)=9`?

Answer 1

Please refer to the Explanation.

Explanation:

`f(x)=(4x^2-7x-2)/(x-2), x!=2`.

`:. f(x)={(4x+1)(x-2)}/(x-2), x!=2`.

`because x!=2, (x-2)!=0," we can divide by "(x-2)" to get,"`

`f(x)=4x+1, x!=2..............(star)`.

Now, we are given that, `|f(x)-9| lt 1/100`.

`:. |(4x+1)-9| lt 1/100," for "x!=2.........[because (star)]`.

`:. |4x-8|=|4(x-2)|=4|x-2| lt 1/100, or,`

`" for "x!=2, 0 lt |x-2| lt 1/400`.

Comparing with `0 lt |x-2| lt delta," clearly, "delta_max=1/400`.

This means that, for a given `epsilon=1/100 gt 0`, we

can always find a `delta=1/4epsilon=1/400` such that,

`0 lt |x-2| lt delta rArr |f(x)-9| lt epsilon`.

By the definition of limit, we conclude that,

`lim_(x to 2) f(x)=9," where, "f(x)=(4x^2-7x-2)/(x-2), x!=2`.

Enjoy Maths.!