What is the limit of (x+7)/(3x+5) as x approaches infinity?
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calendarEducator since 2010
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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
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calendarEducator since 2008
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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 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`
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