Question

Limit x tends to 0 try to solve this

Answer 1

Answer:

1) Depends on n

In fact, the limit is:

• √(1+√2) - √2 ≈ 0.14,    for n = 0;
• 1 / 4√2≈ 0.18,              for n > 0;
• 0,                                for n < 0.

Step-by-step explanation:

If n = 0, then this is constant for x≠0, so the limit is simply that constant value of √(1+√2) - √2 ≈ 0.14.

For n ≠ 0, evaluate as follows.

Two times, multiply numerator and denominator by a "conjugate" to get rid of some square root signs by "difference of squares".

( √( 1 + √( 1 + xⁿ ) ) - √2 ) / xⁿ

         [ first one ]

= ( √( 1 + √( 1 + xⁿ ) ) - √2 )( √( 1 + √( 1 + xⁿ ) ) + √2 ) / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )

         [ difference of squares in numerator ]

= ( ( 1 + √( 1 + xⁿ ) ) - 2 ) / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )

        [ tidy it up ]

= ( √( 1 + xⁿ ) ) - 1 ) / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )

       [ second one ]

= ( √( 1 + xⁿ ) ) - 1 )( √( 1 + xⁿ ) ) + 1 ) / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )( √( 1 + xⁿ ) ) + 1 )

       [ difference of squares in numerator ]

= ( 1 + xⁿ - 1 ) / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )( √( 1 + xⁿ ) ) + 1 )

       [ tidy it up ]

= xⁿ / xⁿ(√( 1 + √( 1 + xⁿ ) ) + √2 )( √( 1 + xⁿ ) ) + 1 )

= 1 / (√( 1 + √( 1 + xⁿ ) ) + √2 )( √( 1 + xⁿ ) ) + 1 )    ... (*)

Now letting x -> 0, we see that for n > 0, this tends to:

 1 / (√( 1 + √( 1 + 0 ) ) + √2 )( √( 1 + 0 ) ) + 1 )

= 1 / (√(1+1) + √2 )( 1 + 1 )

= 1 / ( 2√2 ) ( 2 )

= 1 / 4√2

However, for n < 0, the denominator in (*) tends to ∞, so the limit is 0.