What is the perimeter of a rectangle in which
A= 154 in² and h = 11 in?

0.05

Step-by-step explanation:

We have to find the limit of

as x tends to 100.

If we substitute 100 for x directly we get 0/0 form which is indeterminate.

Let us factorise the denominator suitable

Denominator =

Now we find common factor between numerator and denominator

Since x tends to 100 we can say sqrt x -10 is non zero.

We cancel this andfind

the expression as

Now substitute limits as x =100

Limit =

Limit is 0.05