TS EAMCET · Maths · Limits
If \(\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}=\frac{11}{9}\), then \(\lim _{x \rightarrow a} \frac{x^2+9 x+20}{x^2-x-20}=\)
- A -9
- B -4
- C \(-\frac{1}{4}\)
- D \(-\frac{1}{9}\)
Answer & Solution
Correct Answer
(D) \(-\frac{1}{9}\)
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow 4} \frac{2 x^2+(3+2 a) x+3 a}{x^3-2 x^2-23 x+60}\) [Use L'Hospital] \(\Rightarrow \lim _{x \rightarrow 4} \frac{4 x+2 c t+3}{3 x^2-4 x-23}=\frac{19+2 a}{9}=\frac{11}{9} \Rightarrow a=-4\) Now, \(\lim _{x \rightarrow-4} \frac{x^2+9 x+20}{x^2-x-20}\) [Use…
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