# What is the limit of (x+7)/(3x+5) as x approaches infinity?

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We need to find the limit of the following.

==> lim (x+7) / (3x+5) as x--> inf.

We will divide both numerator and denominator by x.

==> lim ( x+ 7)/x / (3x+5)/x

Now we will simplify.

==> lim ( 1+ 7/x) / (3 + 5/x)

Now we will substitute with x = inf.

==> lim (1+ 7/x) / (3+ 5/x) as x--> inf = ( 1+ 7/inf ) / ( 3+ 5/inf)

But we know that a/ inf = 0

==> lim (x+7)/(3x+5) as x--> inf = (1+ 0)/ (3+ 0) = 1/3

**Then the limit is 1/3.**

The value of lim x--> inf. [(x+7)/(3x+5)] is required.

As x --> inf. , 1/x --> 0

lim x--> inf. [(x+7)/(3x+5)]

=> lim x --> 0[ 1/x + 7)/(3/x + 5)]

=> lim x --> 0[ 1 + 7x)/(3 + 5x)]

substitute x = 0

=> 1/3

**The required limit is 1/3**

The limit `lim_(x-> oo) (x+7)/(3x+5)` has to be determined.

If we substitute `x = oo` in `(x+7)/(3x+5)` we get the indeterminate form `oo/oo` , To determine the limit in this case we can use l'Hospital's rule and substitute the numerator and denominator by their derivatives.

(x +7)' = 1

(3x + 5)' = 3

This gives the limit `lim_(x->oo)1/3`

As the variable x does not figure in `1/3` , this is the required limit.

The limit `lim_(x-> oo) (x+7)/(3x+5) = 1/3`