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lim(x->0) (log(2+x^2)-log(2-x^2))/x=k FI...

`lim_(x->0) (log(2+x^2)-log(2-x^2))/x=k` FInd k value

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`lim_(x->0) (log(2+x^2) - log(2-x^2))/x`
This limit is a `0/0` form. so, we will use `L'`Hospital rule to solve this.
`=>lim_(x->0) (log(2+x^2) - log(2-x^2))/x = lim_(x->0) ((2x)/(2+x^2)-(-2x)/(2-x^2))/1`
When, we put `x = 0`,
`=> lim_(x->0) ((2x)/(2+x^2)-(-2x)/(2-x^2))/1 = (0+0)/1 = 0`
`:. lim_(x->0) (log(2+x^2) - log(2-x^2))/x = 0->(1)`
It is given that ` lim_(x->0) (log(2+x^2) - log(2-x^2))/x = k->(2)`
From (1) and (2),
...
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