How do you find \lim _ { x \rightarrow 0} \frac { x + \sin ( x ) } { x }?

2 Answers
Jun 14, 2017

2

Explanation:

You would substitute the expression:

(x+sinx)/x=x/x+sinx/x=1+sinx/x

Then you will find that

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

Jun 14, 2017

The limit is 2.

Explanation:

lim_(x->0)(x+sinx)/x=0/0

Since it is in the indeterminate form 0/0 you may apply l'Hospital's Rule, that is given f(a) and f(b) is differentiable:

(lim_(x->c)f'(a))/(lim_(x->c)f'(b))=(lim_(x->0)d/dx[x+sinx])/(lim_(x->c)d/dx[x])=(lim_(x->0)(1+cosx))/(lim_(x->c)1)=(1+cos(0))/1=(1+1)/1=2

Thus the limit is 2.