91Ó°ÊÓ

When we talk about evaluating limits, we are asking what value a function approaches as the input gets closer and closer to a certain number, in this case, zero. This doesn't always mean plugging the number directly into the equation, but rather understanding the behavior of the function near that point. To evaluate the limit of a function:

• Identify the function and the point it approaches, such as \( \lim_{h \rightarrow 0}(2h + h^2) \).
• Consider how each term in the function behaves as \( h \) gets closer to the point of interest.
• Some limits may not exist, especially if the function behaves erratically.

In this case, we see that both terms in the expression, \( 2h \) and \( h^2 \), approach zero as \( h \) does. It is crucial to consider each part of the expression to see their contributions, helping determine the overall limit.

Approaching zero means that we are interested in what happens to the values of our function as the variable settles near the point zero. This can tell us much about the continuity of the function and its limiting behavior. Mathematically, as \( h \) approaches zero, the values might not precisely reach zero but will get infinitely close.

• For the term \( 2h \), as \( h \) approaches zero, \( 2h \) also moves towards zero because it is a straightforward linear multiplication.
• For \( h^2 \), as \( h \) approaches zero, the output \( h^2 \) approaches zero even faster. This is because squaring a tiny number makes the result even tinier.

In evaluating the limit, understanding how the function behaves as its inputs near a specified point can reveal a lot about its nature.

The substitution method is one of the common techniques used in evaluating limits. It involves directly substituting the point the variable is approaching into the function if it's possible without causing any undefined behavior. For example, in our problem \( \lim_{h \rightarrow 0}(2h + h^2) \):

• Substitute \( h = 0 \) directly into the expression \( 2h + h^2 \).
• Compute the result: \( 2(0) + 0^2 = 0 \).

In this case, direct substitution works perfectly because neither \( 2h \) nor \( h^2 \) causes an undefined result when \( h \) equals zero. However, remember that not all functions are amenable to the substitution method. It only works well when the function remains defined and does not involve any division by zero or forms that create indeterminate expressions, like \( \frac{0}{0} \). For these, other methods or more advanced techniques might be necessary.