Question

limit x tends to infinite x (log ( 1 + x/2) - log
x/2 =​

Answer 1

Answer:

How to find the limit

limx→+∞(x−x2log(1+1x))

in a elementary way? I can solve with Taylor expansion, but it is placed in the beginning of my calculus book, so I should only use things like:

-Main theorems involving limits, including the limits for x→0, limsinxx, limex−1x, limlog(x+1)x, lim(x+1)p−1x

-x1+x≤log(1+x)≤x or similar inequalities

I cannot use derivatives, Taylor expansion, o(x),O(x) and similar things.

Using the inequality that I have written above and the substitution x=1sint I have been only able to prove that the limit is greater or equal than 0 and smaller or equal than 1.

Answer 2

The value of the limit is 2.

Given:   $\lim_{x \to \infty} x( log (1 + x/2) - log (x/2))$

To Find: The value of the limit.

Solution:

The given limit is,

     $\lim_{x \to \infty} x( log (1 + x/2) - log (x/2))$

  ⇒ $\lim_{x \to \infty} x( log ((1+x/2)x/2)))$                 [ As log A - log B = log A/B ]

  ⇒ $\lim_{x \to \infty} x( log ( 2+x)x)$

  ⇒ $\lim_{x \to \infty} x( log ( 2/x + 1 )$

  ⇒ $\lim_{x \to \infty} log ( 2/x + 1 ) / ( 1/x)$

Putting x = ∞, we get 0/0 form. So, we shall apply L'Hospital's rule here according to which we shall differentiate the numerator and denominator separately.

   ⇒ $\lim_{x \to \infty}$  ((1 / ( 2/x + 1 )) × ( -2 / x² )) / ( -1 / x² )

   ⇒ $\lim_{x \to \infty} 2 / ( 1 + 2/x )$

We know that 1/∞ = 0, so putting x = ∞, we get

   ⇒ 2 / ( 1 + 0 )

   ⇒ 2

Hence, the value of the limit is 2.

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