If `f(x)={(x^2-1)/(x-1);\ \ \ for\ x!=1\ \ \ 2;\ \ \ \ for\ x=1` . Find whether `f(x)` is continuous at `x=1.`

Explanation

`f(x)=(begin{array}{c}`

`
frac{x^{2}-1}{x-1}, x neq 1 `

`
2,` if `x=1`


We see that


(LHL at `x=1`)=

`

lim _{x rightarrow 1^{-}} f(x)=lim _{h rightarrow 0} f(1-h) `

`
lim _{h rightarrow 0} frac{(1-h)^{2}-1}{(1-h)-1}=lim _{h rightarrow 0} frac{1+h^{2}-2 h-1}{1-h-1}=lim _{h rightarrow 0} frac{h^{2}-2 h}{-h}=lim _{h rightarrow 0} frac{h(h-2)}{-h}=lim _{h rightarrow 0}(2-h)=2`


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