Calculate limit with special limits limit sin(x^2+x)/x x-->0

Expert Answers
sciencesolve eNotes educator| Certified Educator

Since the problem requests for you to evaluate the limit using the formula of a special limit, you need to use the following formula of trigonometric remarkable limit, such that:

`lim_(f(x)->0)(sinf(x))/(f(x)) = 1`

You need to notice that the argument of sine function needs to coincide to the denominator of the fraction, hence, for you to use the special limit, you need to divide the function `sin(x^2+x)` by `x^2+x` and multiply it by the same argument, such that:

`lim_(x->0) (sin(x^2+x))/(x^2+x)*(x^2+x)/x`

You need to use the property of distributivity of limit, such that:

`lim_(x->0) (sin(x^2+x))/x = lim_(x->0) (sin(x^2+x))/(x^2+x)*lim_(x->0) (x^2+x)/x`

`lim_(x->0) (sin(x^2+x))/x = 1*lim_(x->0) (x^2+x)/x`

You need to factor out x to numerator, such that:

`lim_(x->0) (x^2+x)/x = lim_(x->0) x(x + 1)/x`

Reducing duplicate factors, yields:

`lim_(x->0) (x^2+x)/x = lim_(x->0) (x + 1) = 0 + 1 = 1`

`lim_(x->0) (sin(x^2+x))/x = 1*1 = 1`

Hence, evaluating the given limit, using the trigonometric special limit `lim_(f(x)->0)(sinf(x))/(f(x)) = 1` , yields `lim_(x->0) (sin(x^2+x))/x = 1` .