Math Questions and Answers

Start Your Free Trial

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

Expert Answers info

james0tucson eNotes educator | Certified Educator

calendarEducator since 2008

write14 answers

starTop subjects are Science, Literature, and Math

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

We can simplify this problem without...

(The entire section contains 76 words.)

Unlock This Answer Now


check Approved by eNotes Editorial


revolution | Student

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.

neela | Student

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.

Is this helpful ?