# Evaluate the limit (3x+5)/(x^2+1), x->+infinity?

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### 4 Answers

limit (3x+5)/(x^2+1) where x--> inf

Since the highest power is x^2, then we divide both functions by x^2

=> lim (3x+5)/lim(x^2+1) = lim x^2(3/x + 5/x^2) /limx^2(1+1/x^2)

==> lim (3/x + 5/x^2) / lim (1+1/x^2) = 0+0/1+0= 0

Then the limit when x--> inf is 0

The limit `lim_(x->oo) (3x+5)/(x^2+1)` has to be determined.

If we substitute `x = oo` in `(3x+5)/(x^2+1)` the numerator and denominator are both equal to 0 which gives a result of the form `oo/oo` . This is an indeterminate form and one that allows the use of l'Hospital's rule to substitute the numerator and denominator by their derivatives.

This makes the limit:

`lim_(x->oo) 3/(2x)`

Now substituting `x = oo` gives the result `3/oo` which is equal to 0.

The required limit `lim_(x->oo) (3x+5)/(x^2+1) = 0`

In order to calculate the limit of a rational function, when x tends to +inf., we'll divide both, numerator and denominator, by the highest power of x, which in this case is x^2.

We'll have:

lim (3x+5)/(x^2+1) = lim (3x+5)/lim (x^2+1)

lim (3x+5)/lim (x^2+1) = lim x^2*(3/x + 5/x^2)/lim x^2*(1 + 1/x^2)

After reducing similar terms, we'll get:

lim (3x+5)/lim (x^2+1) = (0+0)/(1+0)=0

To find the lt (3x+5)/(x^2+1) as x--> infinity.

Soltion:

Here as x--> infinity both numerator and denominator go ifinity, whoch is an indeterminate form of the type: infinity/infinity . Such problems could be solved either by using L' Hospital's rule of differenting both numerator and the denominator and then taking limits. Or we can go dividing term by term numerator and denominator by the highest degree of the variable of the leading term of the denominator and numerator.

Lt (3x+5)/(x^2+1) = Lt(3x/x^2+5/x^2)/(x^2/x^2 +1/x^2) as x--> inf

=Lt {(3*0+1*0)/(1+0)} = 0/1 = 0