Student100 said:I think I stole your name, or you stole mine. ;)
Maybe it would also be helpful if you looked at this way:
When thinking about limits, h is never actually 0. We're interested in what happens as h gets so close to zero we basically treat it as such. Since what you wrote is equivalent to saying [tex] \frac {0}{.000000...1} [/tex] you can conclude that this limit is 0.
If on the other hand you had wrote:
[tex] \lim_{h\rightarrow 0}\frac{(1+h)-1}{h}= 1 [/tex]
Would be the correct interpretation of the limit.
You'll learn more about limits when you discuss the epsilon/delta definition of a limit.
The limit of (1-1)/h as h goes to 0 is undefined. This is because as h approaches 0, the denominator becomes smaller and smaller, approaching 0. When the denominator is 0, the fraction becomes undefined.
The limit of (1-1)/h as h goes to 0 is undefined because when h becomes extremely small, the fraction becomes undefined. This is because dividing by a very small number is equivalent to dividing by 0, which is undefined.
No, the limit of (1-1)/h as h goes to 0 cannot be solved. This is because dividing by 0 is undefined in mathematics, and as h approaches 0, the denominator becomes smaller and smaller, eventually becoming 0.
The limit of (1-1)/h as h goes to 0 cannot be graphically represented. This is because the graph would have an asymptote at h=0, where the function would be undefined. Therefore, the graph would not be continuous and would not accurately represent the limit at h=0.
No, the limit of (1-1)/h as h goes to 0 is not equal to 0. This is because as h approaches 0, the denominator becomes smaller and smaller, causing the fraction to approach infinity. Therefore, the limit is undefined and not equal to 0.