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Evlute [limt(x to 0)(tan^(-1)x)/(x)], wh...

Evlute `[limt_(x to 0)(tan^(-1)x)/(x)]`, where `[*]` re[resemts the greatest integer function.

Text Solution

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See the graphs of y=x and `tan^(-1)x` in the following figure.

For `f(x)=tan^(-1)(x)=1/(1+x^(2))`
`:." "f'(0)=1`
So the curve touches the line y = x at x = 0
For `xgt0,f'(x)lt1`
So the graph of `y=tan^(-1)x` lies below the graph of y = x.
From the graph, when `xto o^(+)`, the graph of y = x is above the graph of `y = tan^(-1)x`
`or" "tan^(-1)xltxrArr(tan^(-1)x)/xlt1rArr[underset(xto0^(+))("lim")(tanx)/x]=0`
`"Thus "[underset(xto0^(+))("lim")(tanx)/x]=0`
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