Question

Help with these questions please!!

Answer 1

Answer:

(1)

option-B

(2)

f(x) is continuous at a=4

Step-by-step explanation:

(1)

we are given

$\lim_{x \to 0} \frac{sin(2x)}{x}$

Since, we are suppose to find limit x-->0

so, we always choose value of x that is  close to 0

At x=-0.03:

$\frac{sin(2\times (-0.03))}{(-0.03)}=1.99880$

At x=-0.02:

$\frac{sin(2\times (-0.02))}{(-0.02)}=1.99947$

At x=-0.01:

$\frac{sin(2\times (-0.01))}{(-0.01)}=1.99987$

At x=0.01:

$\frac{sin(2\times (0.01))}{(0.01)}=1.99987$

At x=0.02:

$\frac{sin(2\times (0.02))}{(0.02)}=1.99947$

At x=0.03:

$\frac{sin(2\times (0.03))}{(0.03)}=1.99880$

(2)

we are given

$f(x)=\frac{x-4}{x+5}$

Since, we have to check continuity at a=4

So, firstly we will find limit value and then functional value

Limit value:

$\lim_{x \to a}  f(x)=\lim_{x \to a}\frac{x-4}{x+5}$

now, we can plug a=4

$\lim_{x \to 4}  f(x)=\lim_{x \to 4}\frac{4-4}{4+5}$

$\lim_{x \to 4}  f(x)=0$

Functional value:

We can plug x=4 into f(x)

$f(4)=\frac{4-4}{4+5}$

$f(4)=0$

So, we can see that

$\lim_{x \to 4}  f(x)=f(4)=0$

So, limit value is equal to function value

so, f(x) is continuous at a=4.............Answer