Question #1f024

2 Answers
Oct 5, 2016

#0#

Explanation:

#(sin(x) - tan(x))/(sin(x)+tan(x))=sin(x)/(sin(x))(1-1/cos(x))/(1+1/cos(x))=#
#(1-1/cos(x))/(1+1/cos(x))=cos(x)/cos(x)(cos(x)-1)/(cos(x)+1)=(cos(x)-1)/(cos(x)+1)#

so

#lim_(x->0)(sin(x) - tan(x))/(sin(x)+tan(x))=lim_(x->0)(cos(x)-1)/(cos(x)+1)=0/2=0#

Oct 5, 2016

#0.#

Explanation:

We use the following Standard Limits :

#lim_(xrarr0)sinx/x=1, and, lim_(xrarr0)tanx/x=1.#

Now, get the Reqd. Limit, we divide by #x#, in #"Nr. and "Dr.# of the

given fun. Thus, we have,

The reqd. Lim. #=lim_(xrarr0) (sinx/x-tanx/x)/(sinx/x+tanx/x)=(1-1)/(1+1)#

#:." The Reqd. Lim."=0#