If `f(x)={e^(1//x),\ \ \ if\ x!=0 1,\ \ \ if\ x=0` . Find whether `f` is continuous at `x=0`

Explanation

`f(x)=e^{frac{1}{x}}, i f x 0 `
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1, i f x=0`

LHL at `x=0)= `
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lim _{x rightarrow 0^{-}} f(x)=lim _{h rightarrow 0} f(0-h)=lim _{h rightarrow 0} f(-h) `
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lim _{h rightarrow 0} e^{frac{-1}{h}}=lim _{h rightarrow 0}(frac{1}{e^{frac{1}{h}}})=frac{1}{lim _{h rightarrow 0} e^{frac{1}{h}}}=0`
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(RHL x=0)= `
`
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