Answer:
[tex]\displaystyle \lim_{x \to 2} \frac{x}{f(x)+1} = +\infty[/tex]
Step-by-step explanation:
From the graph, we can see that:
[tex]\displaystyle \lim_{x \to 2} f(x)= -1[/tex]
From direct substitution, we have that:
[tex]\displaystyle \lim_{x \to 2} \frac{x}{f(x)+1}\Rightarrow \frac{(2)}{-1+1}[/tex]
Evaluate:
[tex]\displaystyle = \frac{2}{0}=\text{Und.}[/tex]
Saying undefined (or unbounded) would be correct.
However, note that as x approaches two, the value of y decreases in order to get to negative one. In other words, our function f will always be greater or equal to negative one (you can also see this from the graph). This means that as x approaches two, f(x) will approach -0.99, -0.999 and -0.9999 and so on until it reaches negative one and then go back up. Importantly, because of this, we can state that:
[tex]\displaystyle \lim_{x \to 2} \frac{x}{f(x)+1}=\frac{(2)}{-1+1} = +\infty[/tex]
This is because for the denominator, the +1 will always be greater than the f(x). This makes this increase towards positive infinity. Note that limits want the values of the function as it approaches it, not at it.
