So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one. 1) lim x → 0 ( e 4 ( 1 − x 2 ) 1 x 2 ( 1 − 2 x ) 1 x ) 1 x I am having problems with this one to make it into one that can be used for L'Hospital's rule, but i cannot get it into 0 0 , there are others options as well, but i am stuck.2) lim n → ∞ ( arcsin ⁡ ( 1 n ) + arcsin ⁡ ( 2 n ) + . . . + arcsin ⁡ ( n n ) n )So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L'Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.

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So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L&#039;Hospital&#039;s rule. I am not sure for the second one. 1)      lim          x      →      0                  (                                    e                          4                                                          (              1              −                              x                                  2                                            )                                                      1                                  x                                      2                                                                                                            (            1            −            2            x            )                                              1              x                                          )                      1        x            I am having problems with this one to make it into one that can be used for L&#039;Hospital&#039;s rule, but i cannot get it into       0    0  , there are others options as well, but i am stuck.2)       lim          n      →      ∞        (            arcsin      ⁡      (              1        n            )      +      arcsin      ⁡      (              2        n            )      +      .      .      .      +      arcsin      ⁡      (              n        n            )        n    )So i thought i could solve this one even if i used comparing test, but i cannot fins a way, but this one is probably not solved with L&#039;Hospital rule, because of n, but i put it in one question, because i found both limits in the same group. I apologize for putting them in the same question if they differ too much. Any help would be appreciated.

Isaiah Haynes · Accepted Answer

Step 1For the first if   x  →  0 then the limit does not exist. To prove this let us observe that  (  1  −      x          2            )          1              /                    x                  2                      →      e          −      1        ,  (  1  −  2  x      )          1              /            x        →      e          −      2      as   x  →  0 and hence  f  (  x  )  =      (                            e                      4                          (        1        −                  x                      2                                    )                      1                          /                                      x                              2                                                                (        1        −        2        x                  )                      1                          /                        x                                )    →      e          5        &amp;gt;  1and   g  (  x  )  =  1      /    x  →  ∞ as   x  →      0          +       so   {  f  (  x  )      }          g      (      x      )        →  ∞ as   x  →      0          +      .Step 2But when   x  →      0          −       then   g  (  x  )  →  −  ∞ and hence   {  f  (  x  )      }          g      (      x      )        →  0 as   x  →      0          −      . It follows that the desired limit does not exist. In the above we have used the following standard limits      lim          x      →      0        (  1  +  a  x      )          1              /            x        =      e          a        ,      a          x        →  ∞   as   x  →  ∞   if   a  &amp;gt;  1The second limit is a Riemann sum and can be evaluated easily (as explained in answer from alans).

yamyekay3 · Accepted Answer

Step 1As regards the first limit, by using      lim          t      →      0        (  1  +  a  t      )          1              /            t        =      lim          s      →      +      ∞        (  1  +  a      /    s      )          s        =      e    a   we have that      lim          x      →      0            (                            e                      4                                                (            1            −                          x                              2                                      )                                              1                              x                                  2                                                                                          (          1          −          2          x          )                                      1            x                                )    =                    e        4            ⋅              e                  −          1                            e              −        2              =      e    5    .So the base of the exponential is eventually greater then 2 (something   &amp;gt;  1 and less than       e    5  ).Step 2If   x  →      0    +   then for   0  &amp;lt;  x  &amp;lt;  r for some   r  &amp;gt;  0,  +  ∞  ←      2          1              /            x        &amp;lt;            (                                    e                          4                                                          (              1              −                              x                                  2                                            )                                                      1                                  x                                      2                                                                                                            (            1            −            2            x            )                                              1              x                                          )                      1        x            If   x  →      0    −   then for   −  r  &amp;lt;  x  &amp;lt;  0 for some   r  &amp;gt;  0,  0  &amp;lt;            (                                    e                          4                                                          (              1              −                              x                                  2                                            )                                                      1                                  x                                      2                                                                                                            (            1            −            2            x            )                                              1              x                                          )                      1        x              &amp;lt;      2          1              /            x        →  0.Therefore you have different behaviours for   x  →      0    −   and   x  →      0    +   and the limit for   x  →  0 does not exist!

So i have two limits that i want to solve, so i will put them together in one question, while both are probably solved with help of L'Hospital's rule. I am not sure for the second one.

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