Question

Find the limit of the function algebraically. lim x → 0 -7 + x/x^2​

Answer 1

The limit of the given function is
`+\infty`.

What is the limit of a function?

The limit of a function explains the behavior of a function as the values of the input(x) gets to a point or infinity.

We are to evaluate the limit of the function
`\lim_(x \to0) ((-7+x)/(x^2))`. Using L'Hospital's rule to evaluate the derivatives of the function, we have:

`f'(x)= (d)/(dx)(-7+x) =1`

`g'(x)= (d)/(dx)(x^2) =2x`

`lim_(x\to 0)(f'(x))/(g'(x))`

`lim_(x\to 0\ \ ) (1)/(2x) = + \infty`

Thus, we can conclude that as 2x approaches 0, the limit of the function goes infinity.

Answer 2

`\lim_(x\to 0)(-7+x)/(x^2)=-\infty`

Explanation:

The given limit problem is

`\lim_(x\to 0)(-7+x)/(x^2)`

Left hand limit of the function is

`LHL=\lim_(x\to 0^-)(-7+x)/(x^2)=\lim_(h\to 0)(-7+(0-h))/((0-h)^2)`

`LHL=\lim_(h\to 0)(-7-h)/(h^2)`

Applying limit, we get

`LHL=-\infty`

Right hand limit of the function is

`RHL=\lim_(x\to 0^+)(-7+x)/(x^2)=\lim_(h\to 0)(-7+(0+h))/((0+h)^2)`

`RHL=\lim_(h\to 0)(-7+h)/(h^2)`

Applying limit, we get

`RHL=-\infty`

Since, LHL=RHL, therefore limit of the function exist at x=0.

Hence,
`\lim_(x\to 0)(-7+x)/(x^2)=-\infty`.