KCET · Maths · Limits
Consider the following statements Statement 1 : \(\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a}\) is 1 (where \(a+b+c \neq 0\) ).
Statement \(2: \lim _{x \rightarrow 2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}\) is \(\frac{1}{4}\).
- A Only statement 2 is true.
- B Only statement \(l\) is true.
- C Both statements 1 and 2 are true.
- D Both statements 1 and 2 are false.
Answer & Solution
Correct Answer
(B) Only statement \(l\) is true.
Step-by-step Solution
Detailed explanation
Statement 1
\(\begin{aligned}
&\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} \\
&=\frac{a \times 1+b \times 1+c}{c \times 1+b \times 1+a}=\frac{a+b+c}{a+b+c}=1
\end{aligned}\)
Statement 1 is true.
Statement 2
\(\begin{aligned}
&=\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} \\
&=\lim _{x \rightarrow-2} \frac{2+x}{x \times 2} \\
&=\lim _{x \rightarrow-2} \frac{1}{x \times 2}=\frac{1}{(-2)(2)}=-\frac{1}{4}
\end{aligned}\)
Statement 2 is false.
\(\begin{aligned}
&\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} \\
&=\frac{a \times 1+b \times 1+c}{c \times 1+b \times 1+a}=\frac{a+b+c}{a+b+c}=1
\end{aligned}\)
Statement 1 is true.
Statement 2
\(\begin{aligned}
&=\lim _{x \rightarrow-2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} \\
&=\lim _{x \rightarrow-2} \frac{2+x}{x \times 2} \\
&=\lim _{x \rightarrow-2} \frac{1}{x \times 2}=\frac{1}{(-2)(2)}=-\frac{1}{4}
\end{aligned}\)
Statement 2 is false.
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