lim x->infinity ((sqrt(x^2+5))-(sqrt(x^2+3)))
lim x->infinity ((sqrt(x^2+5))-(sqrt(x^2+3)))
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The answer is Zero, but I suspect you knew that before you posed the question.
We can simplify this problem without...
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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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