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How do you find the limit of #sqrt(2-x^2)/x# as #x->0^+#?

Answer 1

#lim_(x->0^+) sqrt(2-x^2)/x = +oo#

For #x > 0#:

#sqrt(2-x^2)/x > 0#

Besides the numerator is bounded in the interval #(0,2)# while the denominator is infinitesimal, so the quotient is positive and unbounded and therefore its limit is infinite.

Thus for any #M in (0,+oo)#, if we choose #delta_M# such that:

#delta_M < sqrt(2/(1+M^2))#

then we can see that:

#delta_M^2 (1+M^2) < 2#

#M < sqrt(2-delta_M^2)/delta_M#

Then for #x in (0, delta_M)#

#sqrt(2-x^2)/x > sqrt(2-delta_M^2)/delta_M > M#

Which proves that:

#lim_(x->0^+) sqrt(2-x^2)/x = +oo#

graph{sqrt(2-x^2)/x [-10, 10, -5, 5]}

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Answer 2

Emily Ammerman

To find the limit of sqrt(2-x^2)/x as x approaches 0 from the positive side, we can simplify the expression. By factoring out an x from the denominator, we get x * sqrt(2-x^2)/x. Canceling out the x terms, we are left with sqrt(2-x^2).

Now, as x approaches 0 from the positive side, the value of x^2 becomes very close to 0. Therefore, 2-x^2 approaches 2. Taking the square root of 2, we get sqrt(2).

Hence, the limit of sqrt(2-x^2)/x as x approaches 0 from the positive side is sqrt(2).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

How do you find the limit of #sqrt(2-x^2)/x# as #x-&gt;0^+#?

How do you find the limit of #sqrt(2-x^2)/x# as #x->0^+#?