How do you find the limit of $sinx/(x+sinx)$ as $x->0$ ?

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Answer 1 · 2016-12-08T02:14:41

Use L'Hôpital's rule and evaluate the resulting expression at 0.

Given: $lim_(xto0)sin(x)/(x + sin(x)) = ?$

Because the expression evaluated at 0, is the indeterminate form, $0/0$ , the use of L'Hôpital's rule is warranted.

Compute the derivative of the numerator:

$(d(sin(x)))/dx = cos(x)$

Compute the derivative of the denominator:

$(d(x + sin(x)))/dx = 1 + cos(x)$

Take the limit of the new fraction:

$lim_(xto0)cos(x)/(1 + cos(x)) = cos(0)/(1 + cos(0)) = 1/2$

According to L'Hôpital's rule, the limit of the original function goes to the same value:

$lim_(xto0)sin(x)/(x + sin(x)) = 1/2$

How do you find the limit of sinx/(x+sinx)sinx/(x+sinx) as x->0x->0?