# How is it solving?

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Another way to evaluate this limit is by using L'Hospital rule:

First, rewrite the expression using the relationship `ctgx = 1/(tgx)`

Then, to evaluate `lim_(x ->pi) (x - pi)/(tgx)` , take the derivative of the numerator and denominator. According to the L'Hospital rule, limit of the original ratio equals the limit of the ratio of the derivatives:

`(x-pi)' = 1`

`(tgx)' = sec^2x`

At `x= pi` , `sec^2(pi) = (-1)^2 = 1`

So,

`lim_(x ->pi) (x-pi)/(tgx) = lim_(x ->pi) 1/(sec^2x) = 1/1 = 1`

**The limit in question equals 1.**

You need to evaluate the limit, hence you need to replace `pi` for x in expression, such that:

`lim_(x->pi)(x - pi)ctg x = (pi - pi)ctg pi = 0*oo`

You need to notice that 0*oo is indeterminate, hence, you may re-write the expression using `(cos x)/(sin x) ` for `ctg x,` such that:

`lim_(x->pi)(x - pi)*(cos x)/(sin x)`

You may use the special limit:` lim_(x->0) x/(sin x) = 1` , such that:

`lim_(x->pi)(x - pi)/(sin x)*lim_(x->pi) cos x`

You should use the substitution, such that:

`x -pi = t => x = t + pi`

Since `x->pi` , hence `x - pi -> 0 => t->0`

`lim_(x->pi)(x - pi)/(sin x) = lim_(t->0)(t)/(sin (t + pi))`

You need to notice that `lim_(t->0)(t)/(sin (t + pi)) = 0/0` , hence you may use l'Hospital theorem, such that:

`lim_(t->0)(t)/(sin (t + pi)) = lim_(t->0)(t')/((sin (t + pi))')`

`lim_(t->0)(t')/((sin (t + pi))') = lim_(t->0)1/(cos (t + pi))`

Replacing 0 for t, yields:

`lim_(t->0)1/(cos (t + pi)) = 1/(cos (0 + pi)) = 1/(cos pi) = 1/(-1) = -1`

Hence, replacing the results -1 for `lim_(x->pi)(x - pi)/(sin x)` , yields:

`lim_(x->pi)(x - pi)/(sin x)*lim_(x->pi) cos x = -1*lim_(x->pi) cos x = -1*cos pi = -1*-1 = 1`

**Hence, evaluating the given limit, yields `lim_(x->pi)(x - pi)ctg x = 1.` **