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Evalute [lim(xto0) (sin^(-1)x)/(x)]=1, w...

Evalute `[lim_(xto0) (sin^(-1)x)/(x)]=1`, where`[*]` represets the greatest interger function.

Text Solution

Verified by Experts

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

For `f(x)=sin^(-1)xthereforef'(x)=1/sqrt(1-x^(2))`
f'(0)=1
So the curvetouches the line y=x at x=0.
For `0ltxlt1, f'(x)gt1`
So the graph of y `sin^(-1)x` lies above the graph of y=x.
From the graph, when `xto0^(+)`, the graph of `y=sin^(-1)` is above the graph of y=x
or `sin^(-1)xgtxrArr(sin^(-1)x)/xgt1rArr[underset(xto0^(+))(lim)(sin^(-1)x)/x]=1`
When `x to 0^(-)`, the graph of `y=sin^(-1)x` is below the graph of y = x
or `sin^(-1)xltxrArr(sin^(-1)x)/xgt1("as x is negative")rArr[underset(xto0^(+))(lim)(sin^(-1)x)/x]=1`
Thus, `[underset(xto0^(+))(lim)(sin^(-1)x)/x]=1`
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