Show that the limit leads to an indeterminate form. Then carry out the two-step procedure: Transform the function algebraically and evaluate using continuity.

$\underset{x\to -1}{lim}\frac{{x}^{2}+2x+1}{x+1}$

ka1leE
2021-10-28
Answered

Show that the limit leads to an indeterminate form. Then carry out the two-step procedure: Transform the function algebraically and evaluate using continuity.

$\underset{x\to -1}{lim}\frac{{x}^{2}+2x+1}{x+1}$

You can still ask an expert for help

escumantsu

Answered 2021-10-29
Author has **98** answers

Begin expression,

$\underset{x\to -1}{lim}\frac{{x}^{2}+2x+1}{x+1}$

Substituting x=-1,

$\frac{{(-1)}^{2}+2(-1)+1}{-1+1}=\frac{0}{0}$ Identifical form

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

Since${(a+b)}^{2}={a}^{2}{b}^{2}ab$

$\underset{x\to -1}{lim}\frac{{(x+1)}^{2}}{x+1}$

$\underset{x\to -1}{lim}(x+1)$

$-1+1=0$

Substituting x=-1,

Waiting

Since

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where F(x) is antiderivative, i.e. $F(x{)}^{\prime}=f(x)$.

If so, why I can't find the equality $\int f(x)dx={\int}_{a}^{x}f(t)dt=F(x)$ anywhere? It expresses the relationship between definite and indefinite integral in such a straightfoward way (assuming this equality is true). So it it true and can I use $\int $ and ${\int}_{a}^{x}$ interchangeably?

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