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Find the value of k if \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 1}} = k\) ?
1. 4/3
2. 3
3. 3/2
4. Limit does not exist

1 Answer

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Best answer
Correct Answer - Option 2 : 3

CONCEPT:

If \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\) does not result into indeterminate form, then we use direct substitution in order to find the limits.

The are 7 indeterminate forms which are as follows:

  • \((\frac{0}{0})\)
  • \(\left( {\frac{{ \pm ∞ }}{{ \pm ∞ }}} \right)\)
  • (∞ - ∞)
  • (0 × ∞)
  • 00
  • 1
  • 0

CALCULATION:

Given: \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 1}} = k\)

As we know that, if \(\mathop {\lim }\limits_{x \to a} f\left( x \right)\)does not result into indeterminate form, then we use direct substitution in order to find the limits.

Here, also we can see that \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 1}}\)does not result into any indeterminate form

So, we can substitute x = 1 in the expression \(\frac{{{x^2} + 3x + 2}}{{{x^2} + 1}}\) in order to find the value of k

⇒ \(\mathop {\lim }\limits_{x \to 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 1}} = 3 = k\)

Hence, Option B is the correct answer.

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