You need to substitute `oo ` for x in equation under limit such that:

`lim_(x-gtoo) (tan^(-1)(x/3))/(sin^(-1)(1/x)) = oo/oo`

You need to use l'Hospital's theorem such that:

`lim_(x-gtoo) ((tan^(-1)(x/3))')/((sin^(-1)(1/x))') = lim_(x-gtoo) (1/(3(1 + x^2/9)))/(-(1/(x^2sqrt(1 - x^2/9))))`

`lim_(x-gtoo) (1/(3(1 + x^2/9)))/(-(1/(x^2sqrt(1 - x^2/9)))) = -(1/3)lim_(x-gtoo) (x^2sqrt(1 - x^2/9))/(x^2(1/x^2 + 1/9))`

`-(1/3)lim_(x-gtoo) (x^2sqrt(1...

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You need to substitute `oo ` for x in equation under limit such that:

`lim_(x-gtoo) (tan^(-1)(x/3))/(sin^(-1)(1/x)) = oo/oo`

You need to use l'Hospital's theorem such that:

`lim_(x-gtoo) ((tan^(-1)(x/3))')/((sin^(-1)(1/x))') = lim_(x-gtoo) (1/(3(1 + x^2/9)))/(-(1/(x^2sqrt(1 - x^2/9))))`

`lim_(x-gtoo) (1/(3(1 + x^2/9)))/(-(1/(x^2sqrt(1 - x^2/9)))) = -(1/3)lim_(x-gtoo) (x^2sqrt(1 - x^2/9))/(x^2(1/x^2 + 1/9))`

`-(1/3)lim_(x-gtoo) (x^2sqrt(1 - x^2/9))/(x^2(1/x^2 + 1/9)) = -(1/3)lim_(x-gtoo) (sqrt(1 - x^2/9))/((1/x^2 + 1/9)) = oo`

**Hence, evaluating the limit to the function yields `lim_(x-gtoo) (tan^(-1)(x/3))/(sin^(-1)(1/x)) = oo.` **