Question:medium

Evaluate the Given limit: \(\lim_{x\rightarrow 0}\) \(\frac{ax+b}{cx}\) + 1

Updated On: Jan 21, 2026
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Solution and Explanation

The given limit is

\[ \lim_{x \rightarrow 0} \left( \frac{ax + b}{cx} + 1 \right). \]

Combine the terms inside the bracket by taking a common denominator:

\[ \frac{ax + b}{cx} + 1 = \frac{ax + b}{cx} + \frac{cx}{cx}. \]

\[ = \frac{ax + b + cx}{cx}. \]

\[ = \frac{(a + c)x + b}{cx}. \]

Split the fraction:

\[ \frac{(a + c)x}{cx} + \frac{b}{cx}. \]

\[ = \frac{a + c}{c} + \frac{b}{cx}. \]

Now, consider the limit as \( x \rightarrow 0 \):

\[ \lim_{x \rightarrow 0} \frac{b}{cx}. \]

Since \( c \neq 0 \), as \( x \rightarrow 0 \), \[ \frac{b}{cx} \rightarrow \infty \quad \text{(if } b \neq 0\text{)}. \]

Hence, the given limit does not exist for \( b \neq 0 \).

Therefore,

\[ \boxed{ \lim_{x \rightarrow 0} \left( \frac{ax + b}{cx} + 1 \right) \text{ does not exist (diverges), if } b \neq 0. } \]

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