How do you find the limit of #(x^2+3)/(x^2+4)# as #x->-oo#?

If we look at the graph of #y=(x^2+3)/(x^2+4)# we can see that it is clear that the limit exists, and is approximately #1#

graph{(x^2+3)/(x^2+4) [-10, 10, -2, 2]}

Now, As #x->-oo# then #x^2->oo# ,but #1/x^2->0#

So, it would be better if we could replace #x^2# with #1/x^2#, or #x^-2#

# lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)((x^2+3))/((x^2+4)) * x^-2/x^-2 #

# :. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(x^-2(x^2+3))/(x^-2(x^2+4)) #

# :. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo) (x^-2x^2+3x^-2) / (x^-2x^2+4x^-2)#

# :. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo) (1+3x^-2) / (1+4x^-2)#

And, using #x^-2->0# as #x->-oo# we have;

# lim_(x->-oo)(x^2+3)/(x^2+4) = (1+0) / (1+0) = 1#

Which is completely consistent with the above graph.

We can also use L'Hospital's Rule as the limit is of an indeterminate form #oo/oo#

So By L'Hospital's Rule;

# lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)((x^2+3)')/((x^2+4)') #
# :. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(2x)/(2x) #
# :. lim_(x->-oo)(x^2+3)/(x^2+4) = lim_(x->-oo)(1) #
# :. lim_(x->-oo)(x^2+3)/(x^2+4) = 1 #