Estimate the following limit using graphs or tables.

Estimate the following limit using graphs or tables. In (1 + h) lim h h?0 In (1 + h) lim h h- (Type an integer or decimal rounded to the nearest hundredth as needed.)

The Correct Answer and Explanation is:

To estimate the limit lim⁡h→0ln⁡(1+h)h,\lim_{h \to 0} \frac{\ln(1 + h)}{h},

we can use graphs or tables. But first, let’s understand what this limit represents and how we can estimate it.


Correct Answer:

1.00\boxed{1.00}


Explanation

This limit is a well-known fundamental limit in calculus: lim⁡h→0ln⁡(1+h)h=1.\lim_{h \to 0} \frac{\ln(1 + h)}{h} = 1.

We can understand and estimate this using a table of values or a graph of the function f(h)=ln⁡(1+h)h.f(h) = \frac{\ln(1 + h)}{h}.

Let’s look at some values of hh close to 0:

hhln⁡(1+h)h\frac{\ln(1 + h)}{h}
-0.10.9486
-0.010.9950
-0.0010.9995
0.0011.0005
0.011.0050
0.11.0486

As hh gets closer to 0 from both sides (positive and negative), the values of the function approach 1. This strongly suggests that: lim⁡h→0ln⁡(1+h)h=1.\lim_{h \to 0} \frac{\ln(1 + h)}{h} = 1.


Why Is This Important?

This limit is the derivative of ln⁡(x)\ln(x) at x=1x = 1. In calculus, the derivative is defined as: f′(a)=lim⁡h→0f(a+h)−f(a)h.f'(a) = \lim_{h \to 0} \frac{f(a + h) – f(a)}{h}.

For f(x)=ln⁡(x)f(x) = \ln(x), we get: ln⁡′(1)=lim⁡h→0ln⁡(1+h)−ln⁡(1)h=lim⁡h→0ln⁡(1+h)h.\ln'(1) = \lim_{h \to 0} \frac{\ln(1 + h) – \ln(1)}{h} = \lim_{h \to 0} \frac{\ln(1 + h)}{h}.

So, the value of this limit tells us the rate of change of the natural logarithm at 1, and it is exactly 1. This is a key concept in understanding how logarithmic functions behave near 1.


Final Answer:

1.00\boxed{1.00}

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