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Question: Answered & Verified by Expert
If $f(x)=\left\{\begin{array}{cl}\frac{x^{2}+3 x-10}{x^{2}+2 x-15} & \text { , when } x \neq-5 \\ \text { a }, & \text { when } x=-5\end{array}\right.$

is continuous at $x=-5$, then the value of 'a' will be
MathematicsLimitsBITSATBITSAT 2013
Options:
  • A $3 / 2$
  • B $7 / 8$
  • C $8 / 7$
  • D $2 / 3$
Solution:
2167 Upvotes Verified Answer
The correct answer is: $7 / 8$
$\lim _{x \rightarrow-5} f(x)=\frac{(x-2)(x+5)}{(x+5)(x-3)}=\frac{-7}{-8}=\frac{7}{8}$

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