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For the sine function, when x is close to zero sin(x) = sin(x) and hence when x is close to zero
Thus to evaluate lim h>0 (sinh  sin0)/(h) you need to consider x approaching 0 from the right and x approaching 0 from the left. These give different answers and thus the limit does not exist. I hope this helps,
Hi Mac. You wrote: But this last step is where you erred. That's because sinh is positive when h is a small positive number or a small negative number near zero. But h is positive or negative, so the left limit is 1 and the right limit is +1. The noequations solution: Another way to think of the problem, and this doesn't even involve equations, is to consider that changing x into x is the same as dropping a mirror on the y axis and seeing the "negative" left half of the graph in the mirror. Then you can easily see that at x = 0, the cosine function is smooth, but the sine function has a cusp (corner), so the former is differentiable but the latter is not. Stephen.  


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