Logarithmic and Exponential Limits: Follow the Approach

Approach a point where the logarithm exists

Imagine moving a point along $y=\ln x$ towards $x=2$. Its height approaches $\ln 2$ from either side. The curve has no break there. We express this by writing $$\lim_{x\to2}\ln x=\ln2.$$

A limit asks about nearby inputs, not necessarily the value assigned at the target. For a logarithm at a positive target, these two questions agree because the logarithm is continuous there. More generally, for a fixed base $b>0$, $b\ne1$, and $c>0$, $$\lim_{x\to c}\log_b x=\log_b c.$$

This is a useful first check: if direct substitution gives a valid ordinary value and the expression is continuous there, use it. There is no need to manufacture a complicated method.

Now compare $x\to0$. The real logarithm accepts only positive inputs, so it can approach zero only from the right. As $x\to0^+$, $\ln x$ decreases without bound. There is no finite limit and no real value $\ln0$. A calculator error at zero does not contradict the behaviour of the nearby curve.

Keep three details visible: the function, the target input and the side of approach when the domain requires one. The symbol $x\to c$ means “x approaches c”; it does not instruct you to set every occurrence of x equal to c before inspecting the expression.

A small change reveals the number 1

The expression $\ln(1+h)/h$ is undefined at $h=0$: its numerator and denominator both become zero. But it is defined for nearby $h\ne0$ with $h>-1$. A limit lets us study those nearby values.

For $h=0.1$, the ratio is about 0.9531; for $h=0.01$, it is about 0.9950. From the other side, $h=-0.1$ gives about 1.0536 and $h=-0.01$ gives about 1.0050. Both approaches suggest $$\boxed{\lim_{h\to0}\frac{\ln(1+h)}h=1.}$$

Here is a geometric reason for the result. The natural logarithm $\ln(1+h)$ can be represented by the signed area under $y=1/t$, from $t=1$ to $t=1+h$. For $h>0$, that area lies between rectangles with heights $1/(1+h)$ and 1. After dividing by the positive width h, $$\frac1{1+h}\le\frac{\ln(1+h)}h\le1.$$ For $-1<h<0$, the corresponding bounds are $$1\le\frac{\ln(1+h)}h\le\frac1{1+h}.$$ In each case the two bounds approach 1, so the trapped ratio does too.

The table supports the pattern; the bounds explain why it continues beyond the displayed decimals. The limit is 1, although the original expression has no value at h=0. Use the exploration to approach from both sides, keeping the logarithm’s argument positive.

The matching exponential limit

Now move along $y=a^h$ near $h=0$. Since $a^0=1$, the numerator $a^h-1$ measures the change in height from the point $(0,1)$. Dividing by h gives the slope of the secant joining that point to $(h,a^h)$.

For a fixed $a>0$, the standard result is $$\boxed{\lim_{h\to0}\frac{a^h-1}{h}=\ln a.}$$ In particular, the answer is 1 when $a=e$, positive when $a>1$, and negative when $0<a<1$. This agrees with the increasing or decreasing shape of the exponential curve.

We can connect this limit to the one we already know. For $a\ne1$, put $u=a^h-1$. Then $u\to0$, $1+u=a^h$, and $\ln(1+u)=h\ln a$. Therefore $$\frac{a^h-1}{h}=\ln a\,\frac{u}{\ln(1+u)}\longrightarrow\ln a.$$ The second factor approaches 1 because it is the reciprocal of the natural-logarithm ratio near its nonzero limit.

What if $a=1$? There is no need for that substitution: the numerator is identically zero, so the quotient is zero for every $h\ne0$. Its limit is 0, matching $\ln1$. An exponential base may equal 1, even though a logarithm base may not. Keep those two rules separate.

Keep the scale of the small change

A familiar limit may be hidden by a coefficient. In $\ln(1+3h)/h$, the small change inside the logarithm is $3h$, not h. Match the denominator to that change before using the standard result:

$$\frac{\ln(1+3h)}h=3\,\frac{\ln(1+3h)}{3h}\longrightarrow3.$$ That factor of 3 records how quickly the inner argument moves. Dropping it is the most common error in this kind of limit.

For any fixed real k, $$\lim_{h\to0}\frac{\ln(1+kh)}h=k.$$ When $k\ne0$, substitute $u=kh$; when $k=0$, the quotient is identically zero for nonzero h. For any fixed k, sufficiently small h-values keep $1+kh$ positive.

The same idea works with exponentials. For $a>0$, $$\lim_{h\to0}\frac{a^{kh}-1}{h}=k\ln a.$$ For nonzero k, multiply and divide by k to expose the standard exponential limit. Negative k is allowed: its sign matters.

When both numerator and denominator vanish, do not stop at “0/0”. It is a signal to reorganise the expression, not an answer. Factor each part into a standard small-change ratio and a coefficient. Check that any denominator whose limit you divide by has a nonzero limiting value. This method keeps the reasoning at the level of the two standard limits.

A limit can describe a changing slope

Take two points on $y=\ln x$: one fixed at $(c,\ln c)$, where $c>0$, and another nearby at $(x,\ln x)$. Their secant slope is $$\frac{\ln x-\ln c}{x-c}.$$ As the second point approaches the first, this expression reveals a limiting slope. We can find it using the natural-logarithm limit, without introducing a new differentiation rule.

Put $h=(x-c)/c$. Then $x=c(1+h)$, and $x-c=ch$. Since x and c are positive near the target, the quotient law gives $\ln x-\ln c=\ln(x/c)=\ln(1+h)$. Hence $$\frac{\ln x-\ln c}{x-c}=\frac1c\frac{\ln(1+h)}h\longrightarrow\frac1c.$$

At $c=1$, the limiting slope is 1; at $c=4$, it is $1/4$. The curve is still rising, but it becomes less steep as the positive input grows. This links the number in the limit to something you can see.

For a different valid logarithm base b, divide the whole expression by $\ln b$. Its limiting slope becomes $1/(c\ln b)$, which is negative for $0<b<1$. The Class 12 derivative chapter develops this same idea into a rule for every positive input. Before continuing, make sure you can explain both the algebraic substitution and the meaning of the secant.