# Regarding the calculation of the slope of a tangent.

I have recently had some of my first lessons in calculus. We've learned to use the well-known formula for the slope of a tangent:

$$m_t=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}$$

When working with this formula, I was told that the equation on the left-side should be simplified such that when $$h=0$$, there would be no division by $$0$$.

Obviously, I'm extremely new to these calculus concepts.

I want to know what happens when the equation cannot be simplified and there is a division by $$0$$. That is, there is no solution. What does a situation like this represent? What would it look like graphically?

• Here $h$ is not zero, it tends to zero. Therefore it's perfectly fine to divide by $h$. Sometimes it's not possible to cancel out $h$ but it still may be possible to find the limit. – Vasya Dec 16 '19 at 15:23
• sometimes the limit does not exist; for example, $f(x)=|x|$ – J. W. Tanner Dec 16 '19 at 15:24
• When algebra alone fails to cancel the $h$, we can invoke other techniques that you will soon learn (the "Squeeze" Theorem, L'Hospital's rule, etc) that come at the problem indirectly. "Differential Calculus" is effectively a course devoted to coping with "$0/0$" situations. – Blue Dec 16 '19 at 15:32
• @Blue Hm, it seems I'm just beginning my journey in calculus! Much more to learn in the future and I can't wait! Thank you for making me aware of this. – Sean Xie Dec 16 '19 at 15:39

There is no division by zero, because you're not evaluating the $$\dfrac{f(a+h)-f(a)}{h}$$ at zero. Instead, you're taking a limit, i.e. you're trying to figure out what this expression approaches, as $$h$$ approaches zero.

There are some times when this limit is not defined. For example, let $$f(x)=\sqrt[3]{x}$$ and let $$a=0$$. Then

$$\dfrac{f(a+h)-f(a)}{h}=\dfrac{\sqrt[3]{a+h}-\sqrt[3]{a}}{h}=\dfrac{\sqrt[3]{h}}{h}=\dfrac{1}{\sqrt[3]{h^2}}.$$

And this expression will approach $$\infty$$ as $$h$$ approaches $$0$$.

• It seems my concept of the "limit" was flawed. Thank you for clarifying this. As $h$ "tends to" $0$ and $\frac{1}{h}$ tends to $\infty$ is what I should think of. – Sean Xie Dec 16 '19 at 15:35
• I had the same issue, when I was learning this material. – user729424 Dec 16 '19 at 15:51

There is no graphic distinction between those cases where you can simplify away the $$h$$, and those cases when you can't. The distinction is a purely algebraic one.

For instance, if $$f(x) = x^3$$, then $$\frac{f(x+h) - f(x)}{h} = \frac{(x+h)^3 - x^3}{h} =\\ \frac{3x^2h + 3xh^2+h^3}{h} = 3x^2 + 3xh + h^2$$ (as long as $$h\neq 0$$). So we can simplify away the $$h$$.

On the other hand, for $$g(x) = e^x$$, we have $$\frac{g(x+h) - g(x)}{h} = \frac{e^{x+h} - e^x}{h}\\ = e^x\cdot \frac{e^h - 1}{h}$$ and this can't go any further. You can't get rid of the $$h$$ in the denominator. You just have to know that $$e$$ is defined specifically to be the number that makes the limit of the final fraction (as $$h\to 0$$) equal to $$1$$. And while yes, $$x^3$$ and $$e^x$$ are visually distinct, there is nothing about their visual differences that corresponds directly to the fact that one can be simplified and the other not.