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Q. `lim_(x->oo)` `[(x - 1)/ (x + 3)]^(x + 2)`    = ?

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user8235304 | Student, Grade 11 | Valedictorian

Posted July 2, 2013 at 2:33 PM via web

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Q. `lim_(x->oo)` `[(x - 1)/ (x + 3)]^(x + 2)`    = ?

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Mary Joy Ripalda | High School Teacher | (Level 3) Educator

Posted July 2, 2013 at 4:01 PM (Answer #1)

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`lim_(x->oo)[(x-1)/(x+3)]^(x+2)`

Note that when we plug-in `x=oo` , the result is indeterminate.

 

`lim_(x->oo)[(x-1)/(x+3)]^(x+2)= [(oo-1)/(oo+3)]^(oo+2)=(oo/oo)^oo`  (indeterminate)

 

So to get its limit, we need to take note that a function is equal to y `( y =f(x) )` . And, when a limit of a function as x approaches a certain value does exist, it is equal to y too `(y=lim_(x->c) f(x))` .

Assuming that the limit of the given function does exist, we may re-write our problem as:

`y=lim_(x->oo)[(x-1)/(x+3)]^(x+2)`

Then, remove the variable x in the exponent by taking the natural logarithm of both sides.

`lny=lim_(x->oo)ln[(x-1)/(x+3)]^(x+2)`

At the right side, apply the exponent property of logarithm which is `ln m^a=aln m` .

`lny=lim_(x->oo) (x+2)ln[(x-1)/(x+3)]`

Then, express the right side as a rational function. To do so, re-write (x+2) as 1/(x+2)^(-1).

`lny=lim_(x->oo)1/(x+2)^(-1)*ln[(x-1)/(x+3)]`

`lny=lim_(x->oo)(ln[(x-1)/(x+3)])/(x+2)^(-1)`

Now that we have a rational function at the right side, apply L'Hospital's Rule. So take the derivative of numerator and denominator, separately.

`lny=lim_(x->oo)((ln[(x-1)/(x+3)])')/((x+2)^(-1)')`

`lny=lim_(x->oo)(1/((x-1)/(x+3))*((x+3)*1-(x-1)*1)/(x+3)^2)/(-1(x+2)^(-2)*1)`

Then, simplify.

`lny=lim_(x-oo)((x+3)/(x-1)*(x+3-x+1)/(x+3)^2)/(-(x+2)^(-2))`

`lny=lim_(x->oo)(1/(x-1)*4/(x+3))/(-(x+2)^(-2))`

`lny=lim_(x->oo)4/((x-1)(x+3))*(-(x+2)^2)`

`lny=lim_(x->oo)-(4(x^2+4x+4))/(x^2+2x-3)`

Since the highest exponent of x is 2, factor out x^2 at the top and bottom.

`lny=lim_(x->oo)-(4x^2(1+4/x+4/x^2))/(x^2(1+2/x-3/x^2))`

Cancel the common factor x^2 to simplify further.

`lny=lim_(x->oo)-(4(1+4/x+4/x^2))/(1+2/x-3/x^2)`

Then, to take the limit of the right side, apply the property `lim_(x->oo) (1/x^n)=0` .

`lny=(4(1+0+0))/(1+0-0)`

`lny=-(4*1)/1`

`lny=-4`

And, convert this equation to its equivalent exponential form to isolate the y.

Note that the exponential form of   `ln m = a`  is  `m=e^a` .

`y=e^(-4)`

`y=1/e^4`

Hence, `lim_(x->oo)[(x-1)/(x+3)]^(x+2)=1/e^4` .

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