# What is `lim_(x-> 1) (cos 3x)/(x^2+2x-1)` and `lim_(x->oo)(7x^3+6x^2+1)/((x+1)(2x^2+2)) `

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### 2 Answers

The limit `lim_(x-> 1) (cos 3x)/(x^2+2x-1)` and` lim_(x->oo)(7x^3+6x^2+1)/((x+1)(2x^2+2))` have to be evaluated.

`lim_(x-> 1) (cos 3x)/(x^2+2x-1)`

substitute x = 1

=> `cos 3/2`

`lim_(x->oo)(7x^3+6x^2+1)/((x+1)(2x^2+2))`

=> `lim_(x->oo)(7 + 6/x + 1/x^3)/((1+1/x)(2+2/x^2))`

As x tends to infinity 1/x tends to 0

=> `7/2`

**The limit** `lim_(x-> 1) (cos 3x)/(x^2+2x-1) = cos 3/2` and `lim_(x->oo)(7x^3+6x^2+1)/((x+1)(2x^2+2)) = 7/2`

### User Comments

This limit does not exist.

**Sources:**