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#.