Chapter 14: Q. 6 (page 905)

find the limit $\lim_{x \to - 1} \frac{x^{2} + x + 2}{x^{2} - 9}$

Short Answer

Expert verified

$\lim_{x \to - 1} \frac{x^{2} + x + 2}{x^{2} - 9} = \frac{- 1}{4}$

Step by step solution

Step 1. Given information

$\lim_{x \to - 1} \frac{x^{2} + x + 2}{x^{2} - 9}$

Step 2. Domain of the function

We can find limit of rational function by $\lim_{x \to c} f (x) = f (c)$ if c is in the domain of f(x) .So to find the domain of f(x)= $\frac{x^{2} + x + 2}{x^{2} - 9}$ we have to put denominator =0

$x^{2} - 9 = 0 (x - 3) (x + 3) = 0$

So domain of the f(x) is all real number except 3and -3.

Step 3. limit of the function

As here c= -1 which is in the domain .so ,

$\lim_{x \to - 1} f (x) = f (- 1) = \frac{(- 1)^{2} + (- 1) + 2}{(- 1)^{2} - 9} = \frac{1 - 1 + 2}{1 - 9} = \frac{- 1}{4}$

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find the limit $\lim_{x \to - 1} \frac{x^{2} + x + 2}{x^{2} - 9}$

Short Answer

Step by step solution

Step 1. Given information

Step 2. Domain of the function

Step 3. limit of the function

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find the limitlimx→-1x2+x+2x2-9

Short Answer

Step by step solution

Step 1. Given information

Step 2. Domain of the function

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One App. One Place for Learning.

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find the limit $\lim_{x \to - 1} \frac{x^{2} + x + 2}{x^{2} - 9}$