# lim x^3sin(3/x) / (x^2+1) x...0 without using L" hospitaliam sory idont have math character. lhope help me.

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You need to remember that you may use special limits to evaluate`lim_(x-gt0)x^3sin(3/x) /(x^2+1).`

Hence, using `lim_(x-gt0) (sin x)/x=1` , you may form the`(sin x)/x` type of function inside your function such that:

`lim_(x-gt0) sin(3/x)/(3/x)*(x^3*(3/x))/(x^2+1)`

Notice that the function is divided and multiplied by the argument (`3/x` ) to form the special function `sin(3/x)/(3/x)` .

`lim_(x-gt0) sin(3/x)/(3/x)*lim _(x-gt0)(x^3*(3/x))/(x^2+1)`

Evaluating separately each limit yields:

`lim_(x-gt0) sin(3/x)/(3/x) = 1`

`lim _(x-gt0)(x^3*(3/x))/(x^2+1) = lim _(x-gt0)(3x^2)/(x^2+1)`

Notice that the orders of numerator and denominator are alike, hence the limit is the ratio of leading coefficients such that:

`lim _(x-gt0) (3x^2)/(x^2+1) = 3/1 = 3`

**Hence, evaluating the limit of the function `x^3sin(3/x) /(x^2+1)` if x is heading to 0 yields`lim _(x-gt0) x^3sin(3/x) /(x^2+1) = 3` .**