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



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Posted on (Answer #1)


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:


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


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).



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.



Then, simplify.





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


Cancel the common factor x^2 to simplify further.


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




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` .



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

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