The answer is Zero, but I suspect you knew that before you posed the question.

We can simplify this problem without resorting to calculus. Recall that

lim x->infty(f(x) + g(x)) = lim x->infty(f(x)) + lim x->infty(g(x))

Now it is easy to see that sqrt(x^2+5) approaches infinity with x, and so does sqrt(x^2+3). Since the limits are equal, the difference of the limits is zero.

The answer is obviously zero. Look at this equation:

### lim x->infinity ((sqrt(x^2+5))-(sqrt(x^2+3)))

Both the sqrt of x^2+5 and x^2+3 reaches infernity, so the limits of the equation would cancel each other, becoming ZERO.

An alternate proof using sandwich concept:

To find Limitx->infinity {sqrt(x^2+5)-sqrt(x^2+3)}

Let us look at what is sqrt(x^2+5)-sqrt(x^2+3). Let it be called f(x). Let us rationalise the numerator , by writing in numerator and denominator form.

We see that f(x) > 0 for all x , as the square root of greater positive number is greater than that of a smaller positive number.

f(x)={sqrt(x^2+5)-sqrt(x^2+3)} {sqrt(x^2+5)+{sqrt(x^2+3)}/{sqrt(x^2+5)+sqrt(x^2+3)}

={(x^2+5)-(x^2+3)}/{sqrt(x^2+5)+sqrt(x^2+3)}

=2/{sqr(x^2+5)+sqrt(x^2+3)} < 2/(2x) = 1/x.

Now taking the limits,

Lim x-> infinity f(x) < lim x->infinity (1/x )=0

but f(x)>0 for all x.

Thus 0<f(x)<0. f(x) is sandwiching between zero below and a zero aprroaching value from above as x approaches infininty. Therefore, Lim x->infinity f(x) = 0

**Therefore, Limit x->infinity {sqrt(x^2+5)-sqrt(x^2+3)} = 0.**

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