Evaluate the following limits:
As we need to find
We can directly find the limiting value of a function by putting the value of the variable at which the limiting value is asked if it does not take any indeterminate form (0/0 or ∞/∞ or ∞-∞,1∞ .. etc.)
Let Z =
As it is taking indeterminate form-
∴ we need to take steps to remove this form so that we can get a finite value.
As, Z =
⇒ Z =
Taking log both sides-
⇒ log Z =
⇒ log Z =
{∵ log am = m log a}
Now it gives us a form that can be reduced to
log Z = {adding and subtracting 1 to cos x to get the form}
Dividing numerator and denominator by cos x – 1 to match with form in formula
∴ log Z =
using algebra of limits –
log Z =
∴ A =
Let, cos x - 1 = y
As x→0 ⇒ y→0
∴ A =
Use the formula -
∴ A = 1
Now, B =
∵ cos x – 1 = -2sin2(x/2) and sinx = 2sin(x/2)cos(x/2)
⇒ B =
∴ B = -cot 0 = ∞
∴ B = ∞
Hence,
log Z =
⇒ loge Z = 0
∴ Z = e0 = 1
Hence,