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 ∞-∞, .. etc.)

Let

∴ we need to take steps to remove this form so that we can get a finite value.

Note: While modifying be careful that you don’t introduce any zero terms in the denominator

As Z =

Multiplying numerator and denominator by √(2+cos x) + 1,we have-

Z =

⇒ Z =

{using a² – b² = (a+b)(a-b)}

⇒ Z =

{using basic algebra of limits}

⇒ Z = =

As, 1+cos x = 2cos²(x/2)

∴ Z =

Tip: Similar limit problems involving trigonometric ratios along with algebraic equations are mostly solved using sandwich theorem.

So to solve this problem we need to have a sin term so that we can make use of sandwich theorem.

∵ sin(π/2 – x) = cos x

∴ Z =

As x→π ⇒ π – x → 0

Let y = π – x

Z =

To apply sandwich theorem we have to get the similar form as described below-

∴ Z =

⇒ Z =

Hence,