Answer:
[tex]\lim_{x \to 5^-}f(x)=54\\\lim_{x \to 5^+}f(x)=-17[/tex]
Step-by-step explanation:
So, we have the function:
[tex]f(x)=x^2+5x+4,x<5\\f(x)=8,x=5\\f(x)=-4x+3,\text{ otherwise}[/tex]
And we want to show that this has a jump discontinuity.
As directed, calculate the limit at x=5 from the left and from the right. Thus:
[tex]\lim_{x \to 5^-}f(x)[/tex]
To calculate this, since we're coming from the left, x is less than 5. So, use the first equation:
[tex]\lim_{x \to 5^-}f(x)\\= \lim_{x \to 5^-}(x^2+5x+4)\\[/tex]
Use direct substitution:
[tex]=(5)^2+5(5)+4\\=25+25+4\\=54[/tex]
Thus, as the limit approaches f(5) from the left, the function approaches 54.
Now, calculate the limit at x=5 from the right. Since we're coming from the right, use the third equation:
[tex]\lim_{x \to 5^+}f(x)\\= \lim_{x \to 5^+}(-4x+3)[/tex]
Direct substitution:
[tex](-4(5)+3)\\=-20+3\\=-17[/tex]
Thus, as the limit approaches f(5) from the right, the function approaches -17.
As you can picture, there's a huge gap between y=54 and y=-17. The limits are not equal and both of the limits do exist. Thus, we have jump discontinuity.
