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Answer :
To solve these limits, we need to evaluate each expression by substituting or calculating as necessary. Here is a breakdown of each one:
[tex]\lim_{{x \to 20}} \sqrt{x} \cdot (2x)[/tex]
Substitute [tex]x = 20[/tex] into the expression:
[tex]\sqrt{20} \cdot (2 \times 20) = \sqrt{20} \cdot 40 = 40 \sqrt{20} = 40 \times 2 \sqrt{5} = 80 \sqrt{5}[/tex][tex]\lim_{{x \to 15}} \sqrt{\frac{x+5}{2}}[/tex]
Substitute [tex]x = 15[/tex] into the expression:
[tex]\sqrt{\frac{15+5}{2}} = \sqrt{\frac{20}{2}} = \sqrt{10}[/tex][tex]\lim_{{x \to 1}} (5x^4 - 4x^3 + 3x^2 - 2x + 1)[/tex]
Substitute [tex]x = 1[/tex] into the polynomial:
[tex]5(1)^4 - 4(1)^3 + 3(1)^2 - 2(1) + 1 = 5 - 4 + 3 - 2 + 1 = 3[/tex][tex]\lim_{{x \to 1}} (x^2 + 3x + 4)[/tex]
Substitute [tex]x = 1[/tex] into the polynomial:
[tex](1)^2 + 3(1) + 4 = 1 + 3 + 4 = 8[/tex][tex]\lim_{{x \to \frac{1}{2}}} (x^2 + 3x + 4)[/tex]
Substitute [tex]x = \frac{1}{2}[/tex] into the polynomial:
[tex]\left(\frac{1}{2}\right)^2 + 3\left(\frac{1}{2}\right) + 4 = \frac{1}{4} + \frac{3}{2} + 4 = \frac{1}{4} + \frac{6}{4} + \frac{16}{4} = \frac{23}{4}[/tex][tex]\lim_{{x \to -2}} (x^3 + x^2 - x)[/tex]
Substitute [tex]x = -2[/tex] into the expression:
[tex](-2)^3 + (-2)^2 - (-2) = -8 + 4 + 2 = -2[/tex][tex]\lim_{{x \to 1}} \frac{x^2 + 4x + 5}{x+1}[/tex]
Substitute [tex]x = 1[/tex] into the rational function:
[tex]\frac{1^2 + 4(1) + 5}{1 + 1} = \frac{1 + 4 + 5}{2} = \frac{10}{2} = 5[/tex]
These calculations show how to approach each limit by directly substituting the given value of [tex]x[/tex] when the limit exists. Each solution provides the value that the function approaches as [tex]x[/tex] gets infinitely close to the specified point. This process is often referred to as direct substitution, which works perfectly for continuous functions.
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