Question

Question #1f024

Answer 1

`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`

Answer 2

`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`