Observe the following limit:
$$ \lim_{h\to0}\frac{(2+h)^2-4}{h} $$ In order to find limit of this function we assume $h$ tends to zero. But at last we put $h=0$. Why?
Asked
Active
Viewed 259 times
2
-
1Are you sure the expression is written correct? Cause if it is, you cant just put $0$. More generaly, you put $0$ when the function is continuous – Tair Galili Apr 15 '20 at 15:46
-
In the last step, you do not put the variable $h = 0$, you put the corresponding limit to $0$ (.i.e $\lim_{h\to 0} h = 0$). – achille hui Apr 15 '20 at 15:53
-
Fortunately there are some functions for which limit at a point is same as their value at the point. And this is what we use as a strategy to evaluate limits. Rewrite your expression in a form (including simplification) such that we reach one of those functions whose value is same as their limit and plug $h=0$. Done!! – Paramanand Singh Apr 16 '20 at 01:31
1 Answers
3
In the expression $$ \frac{(2+h)^2 - 4}{h} \ . $$ you cannot simply substitute $h = 0$ to find the limit at $0$, because you can't divide by $0$.
But when $h \ne 0$ simple algebra tells you that expression is equal to $$ 2 + h \ . $$ which is a polynomial in the variable $h$. Polynomials are continuous functions (you have to prove that) so the limit at $0$ is the value at $0$.
Early in your calculus course you probably don't know that about polynomials so the argument you make is something like "when $h$ is near $0$ we can see directly that $2+h$ is near $2$". There is no argument like that for the original expression, before you simplify algebraically.
Ethan Bolker
- 95,224
- 7
- 108
- 199