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Ber [7]
2 years ago
9

Consider the limit of rational function p(x)l q(x)

Mathematics
1 answer:
Helga [31]2 years ago
8 0

#1

\\ \sf\bull\dashrightarrow {\displaystyle{\lim_{x\to c}}}\dfrac{p(x)}{q(x)}=\dfrac{0}{1}=0

  • The limit tends to 0
  • Hence c=0

\\ \sf\bull\dashrightarrow {\displaystyle{\lim_{x\to c}}}\dfrac{p(x)}{q(x)}=\dfrac{1}{1}=1

  • Here limit tends to 1
  • Hence c=1

\\ \sf\bull\dashrightarrow {\displaystyle{\lim_{x\to 0}}}\dfrac{p(x)}{q(x)}=\dfrac{1}{0}=\infty

  • Here x tends to infty.
  • c=infty

For the fourth one limit also tends to zero

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Below in bold.

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<u>Step-by-step explanation:</u>

Convert everything to "sin" and "cos" and then cancel out the common factors.

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\text{Simplify:}\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\div\bigg(\dfrac{sin(x)cos(x)+sin(x)}{cos(x)}\bigg)\\\\\\\text{Multiply by the reciprocal (fraction rules)}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)cos(x)+sin(x)}\bigg)\\\\\\\text{Factor out the common term on the right side denominator}:\\\\\bigg(\dfrac{cos(x)+1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)(cos(x)+1)}\bigg)

\text{Cross out the common factor of (cos(x) + 1) from the top and bottom}:\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times\bigg(\dfrac{cos(x)}{sin(x)}\bigg)\\\\\\\bigg(\dfrac{1}{sin(x)}\bigg)\times cot(x)}\qquad \rightarrow \qquad \dfrac{cot(x)}{sin(x)}

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