# Evaluate the limit_x->+infinity {(x+1)^2010 +(x+2)^2010+(x+3)^2010+........+(x+10)^2010}/(x^2010+10^2010)

*print*Print*list*Cite

### 1 Answer

The existence of the power of 2010 can be very tricky, but this limit is easy to evaluate.

`lim_(x-gtoo) ((x+1)^2010+(x+2)^2010+(x+3)^2010+........+(x+10)^2010)/(x^2010+10^2010)`

Now let's consider `(x+1)^2010`

If we expand this binomal expression,

we should get,

`(x^2010+...............+1)` All the parts after` x^2010 ` will be in less order than it, like `x^2009,x^2008` etc. So no need of writing them, but you have to keep this in mind.

As for `(x+2)^2010` you will get,

`(x+2)^2010 = (x^2010+.............+2^2010)`

and other also in the same manner.

now if we expand this in the limit.

`lim_(x-gtoo) ((x^2010+...+1)+(x^2010+...+2^2010)+(x^2010+...+3^2010)+......+(x^2010+..+10^2010))/(x^2010+10^2010)`

now if we divide both numerator and denominator by x^2010.

Only the first parts of the expansions will be reduced to 1 and other parts will be divided by some order of x.

`lim_(x-gtoo) (((x^2010+...+1)+(x^2010+...+2^2010)+(x^2010+...+3^2010)+......+(x^2010+10^2010))/x^2010)/((x^2010+10^2010)/x^2010)`

`lim_(x-gtoo) ((1+a_1/x+b_1/x^2+...+1/x^2010)+(1+a_2/x+b_2/x^2+...+2^2010/x^2010)+(1+a_3/x+b_3/x^2+...+3^2010/x^2010)+......+(1+a_10/x+b_10/x^2+...+10^2010/x^2010))/(1+10^2010/x^2010)`

now if we apply limit,

`=((1+0+0+..+0)+(1+0+0+..+0)+(1+0+0+..+0)+...+(1+0+0+..+0))/(1+0)`

`=1*10/1` (since you have 10 parts from (x+1) to (x+10)

`=10.`

Therefore,

`lim_(x-gtoo) ((x+1)^2010+(x+2)^2010+(x+3)^2010+......+(x+10)^2010)/(x^2010+10^2010) = 10`