Question

Find lim h→0 f(2+h)-f(2)/h if f(x)=x^2+x+1

Answer 1

Answer:

5

Step-by-step explanation:

$\displaystyle \lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$

$\displaystyle\lim_{h \to 0} \frac{(2+h)^2 + 2 + h + 1 - 2^2 - 2 - 1}{h}$

$\displaystyle\lim_{h \to 0} \frac{4 + 4h + h^2 + h - 4}{h}$

$\displaystyle\lim_{h \to 0} \frac{h^2 + 5h}{h}$

$\displaystyle\lim_{h \to 0} h + 5$

= 5

Answer 2

If f(x) = $x^{2} +x+1$ then the value of $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$ is 5

What is Continuity function?

A continuous function is a function such that a continuous variation of the argument induces a continuous variation of the value of the function

What is Differentiable function?

A differentiable function is a function in one variable in calculus such that its derivative exists at each point in its entire domain

Given,

f(x) = $x^{2} +x+1$

$\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$

$=\lim_{h \to 0} \frac{(2+h)^{2}+(2+h)+1-(2^{2}+2+1)}{h}\\=\lim_{h \to 0} \frac{h^{2}+4h+4+2+h+1-4-2-1 }{h}\\=\lim_{h \to 0} \frac{h^{2}+4h+h }{h}\\=\lim_{h \to 0} \frac{h^{2}+5h}{h}\\=\lim_{h \to 0} h+5\\=5$

Hence, If f(x) = $x^{2} +x+1$ then the value of $\lim_{h \to 0} \frac{f(2+h)-f(2)}{h}$ is 5

Learn more about Continuity function and Differentiable function here

https://brainly.com/question/18882487

#SPJ2