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The substitution method is a fundamental technique used when finding limits in calculus. It involves directly substituting the value that the variable is approaching into the function. This means if you're asked to find \[ \lim_{{y \rightarrow 2}} \frac{y+2}{y^2+5y+6} \]you plug 2 directly into the expression in place of \( y \). In this exercise, we substitute:

• \(y + 2\) becomes \(2 + 2 = 4\)
• The denominator \(y^2 + 5y + 6\) becomes \(4 + 10 + 6 = 20\)

By performing these substitutions, we get \( \frac{4}{20} \), which simplifies to \( \frac{1}{5} \). The substitution method works effectively when the result is not an indeterminate form.

The technique of direct evaluation is practically the simplest when it comes to finding limits. In this approach, after substituting the value in place of the variable, you solve normally as you would for any algebraic expression. Here, after substituting \( y = 2 \), you're left with simplifying a straightforward fraction \( \frac{4}{20} \) which equals \( \frac{1}{5} \). The process is:

• Substitute \( y = 2 \) into both the numerator and the denominator.
• Simplify the resulting numerical expression efficiently.

Direct evaluation is successful here because the expression does not simplify to an indeterminate form like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). This clarity makes direct evaluation an easy-to-understand method provided the function behaves well at the specified point.

Indeterminate forms in limits are special cases where direct substitution doesn't provide a clear answer, making it impossible to determine the limit at first glance. Common indeterminate forms include \( \frac{0}{0} \) and \( \frac{\infty}{\infty} \). These require further analysis or alternative techniques to find the actual limit.In our example of:\[\lim_{{y \rightarrow 2}} \frac{y+2}{y^2+5y+6}\]substituting \( y = 2 \) directly gave us \( \frac{1}{5} \), not an indeterminate form. Hence, the limit was easily determinable by substitution and direct evaluation. However, had the denominator resulted in zero along with the numerator, you'd face an indeterminate form which would require techniques like factoring, simplification, or L'Hôpital's rule to resolve effectively.Understanding indeterminate forms is essential because it tells us when ordinary substitution won't work and when more intricate methods need to be deployed.