How to Find lim x → 0 ( tan ⁡ x x ) 1 x 2

Question

How to Find       lim          x      →      0                  (                        tan          ⁡          x                x            )                      1                  x          2

iberistazi · Accepted Answer

Whenever we have an expression where both base and exponent are variables, it is best to take logs. Thus if L is the desired limit then

\begin{aligned} \log L & = \log {lim_{x \to 0} {(\frac{\tan x}{x})}^{1 / x^{2}}} \\ = lim_{x \to 0} \log {(\frac{\tan x}{x})}^{1 / x^{2}} (via continuity of log) \\ = lim_{x \to 0} \frac{1}{x^{2}} \log (\frac{\tan x}{x}) \\ = lim_{x \to 0} \frac{1}{x^{2}} \cdot \frac{\tan x - x}{x} \cdot \frac{\log (1 + \frac{\tan x - x}{x})}{\frac{\tan x - x}{x}} \\ = lim_{x \to 0} \frac{\tan x - x}{x^{3}} \cdot 1 \\ = lim_{x \to 0} \frac{\sec^{2} x - 1}{3 x^{2}} (via L'Hospital's Rule) \\ = \frac{1}{3} lim_{x \to 0} \frac{\tan^{2} x}{x^{2}} \\ = \frac{1}{3} \end{aligned}

Hence L=e1/3

Isaiah Farrell · Accepted Answer

L&#039;Hospital&#039;s rule is not the alpha and omega of limits computation!As   tan  ⁡  x  =  x  +                    x        3            3        +  o  (      x    3    )            tan      ⁡      x        x    =  1  +            x      2        3    +  o  (      x    2    )  ,    hence        1          x      2        ln  ⁡            (                  tan      ⁡      x        x              )        =      1          x      2        ln  ⁡            (        1  +            x      2        3    +  o  (      x    2    )            )        =      1    3    +  o  (  1  )  →      1    3  so that      lim          x      →      0                  (                  tan      ⁡      x        x                      )                                      1                      x            2                                =            e                      1        3              .

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