# Evaluate the Limit: `lim_(x->0) (x+sin x/x)`

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### 2 Answers

We have to evaluate the limit `lim_(x->0)(x+sinx/x)`

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

substituting x=0 gives us the form `0/0` which is indeterminate. We can use l'Hopital's rule and substitute the numerator and denominator by their derivatives.

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

substituting x=0 gives 1.

**The value of `lim_(x->0)(x + sin x/x) = 1` **

You can distribute the limit to each member of the sum.

`lim_(x->0)` `(x + (sin x)/x) = lim_(x-gt0)x + lim_(x-gt0) (sin x)/x`

Use the formula `lim_(u_n-gt0) (sin u_n)/(u_n)=1`

`lim_(x-gt0) (x + (sin x)/x)= 0 + 1 = 1`

**ANSWER: The limit of the sum `(x + (sin x)/x)` is 1, if x->0.**