Question

evalute the limit . Lim ( 1-cosπx)/(tan²πx) X->0

Answer 1

Step-by-step explanation:

First determining the form i.e. 0/0.

Now replace x by 1+h, so that x->(1+h) Hence h->0.{“->” symbolizes “Tends to”} so, the limit becomes as lim(h->0)[(1+cos(π+πh)/tan^2(π+πh)]}.

Doing this makes it easier to apply standard limit results.

Now, as h->0 ,πh is an acute angle. Therefore on applying Trigonometry we can say that cos(π+πh) becomes -cos(πh) and {tan(π+πh)}^2 becomes {tan(πh)}^2.

Hence the limit now becomes as, lim(h->0)[1-cos(πh)/{tan(πh)^2}]

Now multiply and divide by (πh)^2.

This makes it as lim(h->0)[1-cos(πh)(πh)^2/{tan(πh)^2}(πh)^2]

Now separate the limit as lim(h->0)[1-cos(πh)/(πh)^2]*

lim(h->0){(πh)^2/{tan(πh)^2}

1.) lim(h->0)[1-cos(πh)/(πh)^2=1/2 (Standard Result)

2.) lim(h->0){(πh)^2/{tan(πh)^2}=1 (Standard Result)

Hence the over all value of limit become as (1/2)*1=1/2

So, lim x~1 (1+ Cos πx) ÷ tan^2πx = 1/2.

Note: This might seem to be a lengthy and tricky process and there is an alternative to it which is short direct and easy known as L Hopitals Rule

In case you don’t know about it. It says that lim x->0 f(x)/g(x) = lim x->0 f’(x)/g’(x)

But can be used in 0/0(like one in this question) or infinity/infinity in determinant form only.

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