Question

Question #6dd46

Answer 1

`lim_(x->0)x^2/(1-cos(x)) = 2`

Explanation:

We will make use of some algebra,the well known limit `lim_(x->0)sin(x)/x = 1`

and the following:

• if`f(x)` is continuous, then `lim_(x->a)f(x) = f(lim_(x->a)x)`
• if `f(x)` and `g(x)` have finite limits at `a`, then `lim_(x->a)f(x)g(x) = lim_(x->a)f(x)*lim_(x->a)g(x)`

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`lim_(x->0)x^2/(1-cos(x))`

`=lim_(x->0)(x^2(1+cos(x)))/((1-cos(x))(1+cos(x)))`

`=lim_(x->0)x^2/(1-cos^2(x))(1+cos(x))`

`=lim_(x->0)x^2/sin^2(x)(1+cos(x))`

`=lim_(x->0)(x/sin(x))^2(1+cos(x))`

`=lim_(x->0)(sin(x)/x)^(-2)(1+cos(x))`

`=lim_(x->0)(sin(x)/x)^(-2) * lim_(x->0)(1+cos(x))`

`=(lim_(x->0)sin(x)/x)^(-2) * lim_(x->0)(1+cos(x))`

`=1^(-2)(1+cos(0))`

`=2`

Answer 2

`2`.

Explanation:

We use the Trigo. Identity `: 1-cos2theta=2sin^2theta`.

Reqd. Lim. `=lim_(xrarr0) x^2/(1-cosx)`

`=lim_(xrarr0)x^2/(2sin^2(x/2)`

`=lim_(xrarr0) (2(x^2/4))/(sin^2(x/2)`

`=2[lim_(xrarr0){(x/2)/(sin(x/2))}^2]`

`=2[lim_(xrarr0)(x/2)/(sin(x/2))]^2`

`=2*(1)^2`

`=2`, as respected sente has derived!.