Answer:
[tex]\lim_{x\to 0}\dfrac{-7+x}{x^2}=-\infty[/tex]
Step-by-step explanation:
The given limit problem is
[tex]\lim_{x\to 0}\dfrac{-7+x}{x^2}[/tex]
Left hand limit of the function is
[tex]LHL=\lim_{x\to 0^-}\dfrac{-7+x}{x^2}=\lim_{h\to 0}\dfrac{-7+(0-h)}{(0-h)^2}[/tex]
[tex]LHL=\lim_{h\to 0}\dfrac{-7-h}{h^2}[/tex]
Applying limit, we get
[tex]LHL=-\infty[/tex]
Right hand limit of the function is
[tex]RHL=\lim_{x\to 0^+}\dfrac{-7+x}{x^2}=\lim_{h\to 0}\dfrac{-7+(0+h)}{(0+h)^2}[/tex]
[tex]RHL=\lim_{h\to 0}\dfrac{-7+h}{h^2}[/tex]
Applying limit, we get
[tex]RHL=-\infty[/tex]
Since, LHL=RHL, therefore limit of the function exist at x=0.
Hence, [tex]\lim_{x\to 0}\dfrac{-7+x}{x^2}=-\infty[/tex].
