indeterminate limit: lim, x->-2, [(x^2-3)/(x^2-x-5)] ^ (3/(x^2+2x))

[newuser]

limit [(x^2-3)/(x^2-x-5)] ^{(3/(x^2+2x))}
x->-2


It forms an indeterminate form 1^infinity. so I used exponential with logarithm and got


limit (3/(x^2+2x)) Ln [(x^2-3)/(x^2-x-5)]
e^ x->-2

My question is how to go next without l´hopital rule? Thanks

 

[newuser]

Thanks. No need to solve now. I got it. and the answer is e^-1.5. Thank you so much.

limit [(x^2-3)/(x^2-x-5)] ^{(3/(x^2+2x))}
x->-2


It forms an indeterminate form 1^infinity. so I used exponential with logarithm and got


limit (3/(x^2+2x)) Ln [(x^2-3)/(x^2-x-5)]
e^ x->-2

My question is how to go next without l´hopital rule? Thanks

Thanks. No need to solve now. I tried and got it. The answer is e^-1.5. Thank you so much.