Question

How do you find `lim (t^3-6t^2+4)/(2t^4+t^3-5)` as `t->oo`?

Answer 1

0

Explanation:

Finding the limit of a function is basically just a way to find out what value we get closer and closer to as we approach a certain number.

Finding the limit at infinity is no different. We should establish a couple of rules before we start.

`...........................barul(| color(white)(----)"Rules"color(white)(----)|).........................`

`color(white)(----)*(oo)/(n) = oocolor(white)(aaaa)"Where n is any integer"`

`color(white)(----)*(n)/(oo) = 0`
`.................................................................................................`

Knowing this, we can go ahead and approach the problem as follows:

`barul"|Step 1|"`

The first thing when taking the limit of a rational function is that we should focus our attention on the denominator. There, we must look at the highest power of the polynomial, which in our case is `color(red)[t^4`.

• `lim_(trarroo)(t^3 - 6t^2 + 4)/(2color(red)barul|t^4| + t^3 - 5)`

`barul"|Step 2|"`

Next, we will take `color(red)[t^4` and divide it by every term in both the numerator AND the denominator. Doing so we get the following:

• `lim_(trarroo) ((t^3/color(red)[t^4]-"6t"^2/color(red)[t^4]+4/color(red)[t^4]))/((("2t"^4)/color(red)[t^4]+t^3/color(red)[t^4]-5/color(red)[t^4]))->lim_(trarroo) ((1/t-6/t^2+4/t^4))/((2+1/t-5/t^4))`

`barul"|Step 3|"`

So, in our final step, we look at our rules that we noted above and simplify. Applying our rules, we get the following answer:

• `lim_(trarroo)((stackrelcolor(blue)"0"cancel(1/t)-stackrelcolor(blue)"0"cancel(6/t^2)+stackrelcolor(blue)"0"cancel(4/t^4)))/((2+stackrelcolor(blue)"0"cancel(1/t)-stackrelcolor(blue)"0"cancel(5/t^4)))`
  `color(white)(aaa)`

• `lim_(trarroo)(color(blue)0-color(blue)0+color(blue)0)/(2+color(blue)0-color(blue)0)`
  `color(white)(aaa)`

• `lim_(trarroo)(0)/(2)`
  `color(white)(aaa)`

• `color(magenta)[ 0`

`"Answer":color(magenta)(0`