-1

enter image description here

I got $(16-16)/0$. Am I wrong to have this conclusion, or am I correct and the answer somehow isn't "DNE" but instead $0$?

Blue
  • 75,673
dagajack
  • 27
  • 1
    you can try plotting the expression. – qwr Apr 24 '20 at 03:12
  • 2
    How about simplifying the expression? You should get something with a factor of h in the numerator, which you can then cancel with the denominator. – The Chaz 2.0 Apr 24 '20 at 03:13
  • Consider a related problem: $\lim_{h\rightarrow 0} \frac{6h}{7h}$. Clearly the answer is $6/7$, even though taking the limits separately in the numerator and denominator gives $0/0$. Getting $0/0$ just means that we cannot find the limit by separately taking limits of the numerator and denominator, we must find another method (which often involves simplifying the expression as suggested in the above comment). – Michael Apr 24 '20 at 03:22
  • 2
    Do you know what "limit" means? $\lim_{x\to a} f(x) \ne f(a)$. That ISN"T how you do limits (unless you can prove $f(x)$ is continuous at $x=a$ (and if $f(a)$ is not defined then $f(x)$ is NOT continuous at $x=a$ and $\lim_{x\to a}f(x) \ne f(a)$. ..... Review what $\lim_{x\to a}f(x)$ actually means. – fleablood Apr 24 '20 at 03:22
  • "Am I wrong to have this conclusion" Yes. "or am I correct" you are not "and the answer somehow isn't DNE" it isn't DNE. "but instead 0" The answer is not $0$. Notice if $h = 0.1$ then $\frac {(-4+h)^2-16}h=\frac {(-3.9)^2 - 16}{0.1} = \frac {-0.79}{0.1} = -7.9$. And if $h=0.01$ then $\frac {(-4+h)^2-16}h=\frac {(-3.99)^2 - 16}{0.01} = \frac {-0.0799}{0.01} = -7.99$.... doesn't look like $\lim_{h\to 0}\frac {(-4+h)^2-16}h=0$ – fleablood Apr 24 '20 at 03:28
  • Not $h\ne 0$. $h$ never is equal to $0$. $h > 0$ and $h$ is always greater than $0$. You do not plug in $h=0$. You have to do something entirely different. – fleablood Apr 24 '20 at 03:31
  • 1
    The expression "$0/0$" is meaningless in algebra. Importantly, however, all of differential calculus is about assigning meaning to expressions that ostensibly evaluate to "$0/0$"! The notion of limit doesn't evaluate an expression "at a point"; rather, it evaluates the expression near a point ... and nearer that point ... and nearer still ... and determines whether the expression is suggesting a value at the point that it simply won't let us evaluate by direct substitution. I like to think this answer describes the phenomenon nicely. – Blue Apr 24 '20 at 03:45
  • It should be noted that $ \lim_{x\to a}f(x) $ is defined keeping $x\ne a$ in mind. We consider a neibourhood of $a$ and not the value at $a$. Even if the value of $f$ at $a$ is not defined, even then limit of $f$at $a$ may exist. Therefore, you should not jump to conclusion by finding value at $h=0$ in your question. You should analyze the neibourhood of $0$. Checkout the comment by Mr. @Michael for example. – Koro Apr 24 '20 at 06:59

1 Answers1

1

You have to break it like this: $$\begin{align}\require{cancel} \lim_{h\rightarrow0} \frac {(-4+h)^2-16}{h}&=\lim_{h\rightarrow0} \frac {(h^2-8.h+16)-16}{h} \\&=\lim_{h\rightarrow0} \frac {h^2-8.h}{h}\\&=\lim_{h\rightarrow0} \frac {\cancel {h}.(h-8)}{\cancel {h}},\text{as $h\rightarrow0$ doesn't mean $h=0$.}\\&=\lim_{h\rightarrow0}(h-8)\\&=-8 \end{align}$$ So neither $0$ is the answer.

Tamas Kanti Garai
  • 304