Please evaluate the limit of the following for me:

1. Lim of x^2/sinx as x approaches to 0.

2. Lim of sinax/tanbx as x approaches to 0.

And also:

3.Lim of sin^2x/tan^2x as x approaches to pi/2.

4. Lim of 2y-pi/cosy as y approaches to pi/2.

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1. You should use the remarcable limit `lim_(x-gt0) x/(sin x) = 1` such that:

`lim_(x-gt0) (x^2)/(sin x) = lim_(x-gt0) x/(sin x)* lim_(x-gt0) x`

`lim_(x-gt0) (x^2)/(sin x) = 1*0 = 0`

If you substitute 0 for x right away, you will see that you will obtain an indetermination such that:

`lim_(x-gt0) (x^2)/(sin x) = 0/0`

If you don't remember the remarcable limit, you also may use l'Hospital's theorem such that:

`lim_(x-gt0) (x^2)/(sin x) = lim_(x-gt0) ((x^2)')/((sin x)') `

`lim_(x-gt0) ((x^2)')/((sin x)') = lim_(x-gt0) (2x)/(cos x) = 0/1 = 0`

**Hence, evaluating the limit yields** `lim_(x-gt0) (x^2)/(sin x) = 0.`

2. You should use the remarcable limits `lim_(x-gt0) (sin ax)/(ax) = 1` and `lim_(x-gt0) (tan bx) /(bx) = 1` such that:

`lim_(x-gt0) (sin ax) / (tan bx) = lim_(x-gt0) (sin ax) /(ax)*lim_(x-gt0) (bx)/(tan bx)*lim_(x-gt0) (ax)/(bx)`

`lim_(x-gt0) (sin ax) / (tan bx) = 1*1*(a/b) = a/b`

**Hence, evaluating the limit yields** `lim_(x-gt0) (sin ax)/(tan bx) =a/b` .

3. You need to evaluate `lim_(x-gtpi/2) (sin^2 x)/(tan^2 x),` hence, you should substitute `pi/2` for x such that:

`lim_(x-gtpi/2) (sin^2 x)/(tan^2 x) = (sin^2(pi/2))/(tan^2(pi/2)) = 1/oo = 0`

**Hence, evaluating the limit yields `lim_(x-gtpi/2) (sin^2 x)/(tan^2 x) = 0` .**

4. You need to evaluate `lim_(y-gtpi/2) (2y-pi)/(cosy) = (2pi/2 - pi)/(cos pi/2)` `lim_(y-gtpi/2) (2y-pi)/(cosy) = 0/0`

You may use l'Hospital's theorem such that:

`lim_(y-gtpi/2) (2y-pi)/(cosy) = lim_(y-gtpi/2) ((2y-pi)')/((cosy)') `

`lim_(y-gtpi/2) ((2y-pi)')/((cosy)') = lim_(y-gtpi/2) 2/(-sin y)`

`lim_(y-gtpi/2) 2/(-sin y) = 2/(-sin pi/2) = 2/(-1)`

`lim_(y-gtpi/2) (2y-pi)/(cosy) = -2`

**Hence, evaluating the limit`lim_(y-gtpi/2) (2y-pi)/(cosy) = -2.` **

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