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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.
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