Respuesta :

Answer:

[tex]\lim_{x\to 1^+} f(x)=5[/tex]

Step-by-step explanation:

So we have the function:

[tex]f(x) = \left\{ \begin{array}{ll} x+2 & \quad x \leq 1 \\ x^2+2x+2 & \quad x > 1 \end{array} \right.[/tex]

And we want to evaluate:

[tex]\lim_{x\to 1^+} f(x)[/tex]

In other words, we want to evaluate the limit for f(x) as x approaches 1 from the right.

Since our x is approaching from the right, this means that we need to use the second equation. This is because the second equation is defined for all values greater than 1. So, if we approach from the right, we are using all values greater than 1. Therefore:

[tex]\lim_{x\to 1^+} f(x)[/tex]

Replace f(x) with the second equation:

[tex]=\lim_{x\to 1^+} (x^2+2x+2)[/tex]

Evaluate. Use direct substitution:

[tex]=(1)^2+2(1)+2[/tex]

Square:

[tex]=1+2(1)+2[/tex]

Multiply:

[tex]=1+2+2[/tex]

Add:

[tex]=5[/tex]

Therefore:

[tex]\lim_{x\to 1^+} f(x)=5[/tex]

And we're done!

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