A limit without invoking L'Hopital: lim x → 0 x cos ⁡ x − sin ⁡ x x 2

Question

A limit without invoking L&#039;Hopital:       lim          x      →      0                  x      cos      ⁡      x      −      sin      ⁡      x              x      2

Elianna Douglas · Accepted Answer

From the geometric proof of             sin      ⁡      x        x    →  1 as   x  →  0 we know   cos  ⁡  x  &amp;lt;            sin      ⁡      x        x    &amp;lt;  1 near 0. Since             x      cos      ⁡      x      −      sin      ⁡      x              x      2        =            cos      ⁡      x      −                        sin          ⁡          x                x              x  , we see that  0  =      lim          x      →      0                  cos      ⁡      x      −      1        x    ≤  ℓ  ≤      lim          x      →      0                  cos      ⁡      x      −      cos      ⁡      x        x    =  0  ,,so   ℓ  =  0.

tr2os8x · Accepted Answer

We have,      lim          x      →      0                          x        cos        ⁡        x        −        sin        ⁡        x                    x        2              =      lim          x      →      0                          cos        ⁡        x        −        1            x        +      lim          x      →      0                          x        −        sin        ⁡        x                    x        2              =  −  2      lim          x      →      0                                    sin          2                ⁡                  (                      x            2                    )                    x        +      lim          x      →      0                          x        −        sin        ⁡        x                    x        2            The first limit is zero since             lim              x        →        0                                      sin          ⁡          x                x              =    1  , and,  0  ≤      lim          x      →      0                          x        −        sin        ⁡        x                    x        2              ≤      lim          x      →      0                          tan        ⁡        x        −        sin        ⁡        x                    x        2            But,      lim          x      →      0                          tan        ⁡        x        −        sin        ⁡        x                    x        2              =      lim          x      →      0               (    sin    ⁡    x    ×                            1          −          cos          ⁡          x                                      x            2                    cos          ⁡          x                      )    =      lim          x      →      0                          1        −        cos        ⁡        x            x        =  0Thus, by the Squeeze Theorem,      lim          x      →      0                          x        cos        ⁡        x        −        sin        ⁡        x                    x        2              =  0

A limit without invoking L'Hopital: <munder> <mo movablelimits="true" form="prefix">lim

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