limit x----> infinity ((x-5)/(x-2))^(x-3) is ?

`lim_(x->oo)((x-5)/(x-2))^(x-3)`

`=lim_(x->oo)((x-5)/(x-2))^x*((x-5)/(x-2))^-3`

Since `lim_(x->a)[f(x)*g(x)]=lim_(x->a)f(x)*lim_(x->a)g(x)`

`=lim_(x->oo)((x-5)/(x-2))^x*lim_(x->oo)((x-5)/(x-2))^-3`

Now let's evaluate the limits of both the functions separately,

`lim_(x->oo)((x-5)/(x-2))^-3`

`=lim_(x->oo)((x(1-5/x))/(x(1-2/x)))^-3`

`=lim_(x->oo)((1-5/x)/(1-2/x))^-3`

apply infinity property,

`=((1-0)/(1-0))^-3`

=1

Now,`lim_(x->oo)((x-5)/(x-2))^x`

`=lim_(x->oo)e^(xln((x-5)/(x-2)))`

Now apply the limit chain rule,

Now evaluate `lim_(x->oo)xln((x-5)/(x-2))`

`=lim_(x->oo)ln((x-5)/(x-2))/(1/x)`

Now let's apply the L'Hospital's rule to evaluate the limit,

Test L'Hospital condition=0/0

`=lim_(x->oo)(d/dx(ln((x-5)/(x-2))))/(d/dx(1/x))`

`=lim_(x->oo)(1/((x-5)/(x-2))d/dx((x-5)/(x-2)))/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*((x-2)d/dx(x-5)-(x-5)d/dx(x-2))/(x-2)^2)/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*((x-2)-(x-5))/(x-2)^2)/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*(3/(x-2)^2))/(-1/x^2)`

`=lim_(x->oo)(-3x^2)/((x-2)(x-5))`

`=lim_(x->oo)(-3x^2)/(x^2-7x+10)`

Now again apply L'Hospital's rule,

`=lim_(x->oo)(d/dx(-3x^2))/(d/dx(x^2-7x+10))`

`=lim_(x->oo)(-3*2x)/(2x-7)`

`=lim_(x->oo)(-6x)/(2x-7)`

Now again apply L'Hospital's rule,

`=lim_(x->oo)(d/dx(-6x))/(d/dx(2x-7))`

`=lim_(x->oo)(-6/2)`

`=lim_(x->oo)-3`

=-3

`:.lim_(x->oo)e^(xln((x-5)/(x-2)))=e^-3=1/e^3`

`:.lim_(x->oo)((x-5)/(x-2))^(x-3)=1/e^3*1`

`=1/e^3`

`lim_(x->oo)((x-5)/(x-2))^(x-3)`

`=lim_(x->oo)((x-5)/(x-2))^x*((x-5)/(x-2))^-3`

Since `lim_(x->a)[f(x)*g(x)]=lim_(x->a)f(x)*lim_(x->a)g(x)`

`=lim_(x->oo)((x-5)/(x-2))^x*lim_(x->oo)((x-5)/(x-2))^-3`

Now let's evaluate the limits of both the functions separately,

`lim_(x->oo)((x-5)/(x-2))^-3`

`=lim_(x->oo)((x(1-5/x))/(x(1-2/x)))^-3`

`=lim_(x->oo)((1-5/x)/(1-2/x))^-3`

apply infinity property,

`=((1-0)/(1-0))^-3`

=1

Now,`lim_(x->oo)((x-5)/(x-2))^x`

`=lim_(x->oo)e^(xln((x-5)/(x-2)))`

Now apply the limit chain rule,

Now evaluate `lim_(x->oo)xln((x-5)/(x-2))`

`=lim_(x->oo)ln((x-5)/(x-2))/(1/x)`

Now let's apply the L'Hospital's rule to evaluate the limit,

Test L'Hospital condition=0/0

`=lim_(x->oo)(d/dx(ln((x-5)/(x-2))))/(d/dx(1/x))`

`=lim_(x->oo)(1/((x-5)/(x-2))d/dx((x-5)/(x-2)))/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*((x-2)d/dx(x-5)-(x-5)d/dx(x-2))/(x-2)^2)/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*((x-2)-(x-5))/(x-2)^2)/(-1/x^2)`

`=lim_(x->oo)(((x-2)/(x-5))*(3/(x-2)^2))/(-1/x^2)`

`=lim_(x->oo)(-3x^2)/((x-2)(x-5))`

`=lim_(x->oo)(-3x^2)/(x^2-7x+10)`

Now again apply L'Hospital's rule,

`=lim_(x->oo)(d/dx(-3x^2))/(d/dx(x^2-7x+10))`

`=lim_(x->oo)(-3*2x)/(2x-7)`

`=lim_(x->oo)(-6x)/(2x-7)`

Now again apply L'Hospital's rule,

`=lim_(x->oo)(d/dx(-6x))/(d/dx(2x-7))`

`=lim_(x->oo)(-6/2)`

`=lim_(x->oo)-3`

=-3

`:.lim_(x->oo)e^(xln((x-5)/(x-2)))=e^-3=1/e^3`

`:.lim_(x->oo)((x-5)/(x-2))^(x-3)=1/e^3*1`

`=1/e^3`