5.2. Find the value of $$ c $$ that will make $$ f $$ continuous at $$ x=0 $$ $$ f(x)=\left\{\begin{array}{lll} x \cot 5 x & \text { if } & x

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5.2. Find the value of $$ c $$ that will make $$ f $$ continuous at $$ x=0 $$ $$ f(x)=\left\{\begin{array}{lll} x \cot 5 x & \text { if } & x<0 \ x^{2}+c & \text { if } & x \geq 0 \end{array}\right. $$

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We need $$ f $$ to be continuous at $$ x=0 $$. The function is defined as
$$
f(x)=
\begin{cases}
x \cot(5x), & \text{if } x < 0, \
x^{2}+c, & \text{if } x \ge 0.
\end{cases}
$$
For continuity at $$ x=0 $$, the limit from the left must equal the function value at $$ x=0 $$. That is, we require
$$
\lim_{x\to 0^-} x \cot(5x) = f(0).
$$
Since $$ f(0) = 0^2 + c = c $$, we must have
$$
\lim_{x\to 0^-} x \cot(5x) = c.
$$

**Step 1.** Express $$ \cot(5x) $$ in terms of sine and cosine:
$$
\cot(5x) = \frac{\cos(5x)}{\sin(5x)}.
$$
Thus,
$$
x \cot(5x) = \frac{x\cos(5x)}{\sin(5x)}.
$$

**Step 2.** Compute the limit as $$ x \to 0^- $$. Since both sine and cosine are continuous and $$\cos(5x) \to \cos(0) = 1$$, we have
$$
\lim_{x\to 0} \frac{x\cos(5x)}{\sin(5x)} = \lim_{x\to 0} \frac{x}{\sin(5x)}.
$$
Notice that the cosine function does not affect the limit because it approaches $$1$$.

**Step 3.** Use the substitution $$ t = 5x $$:
- When $$ x \to 0 $$, then $$ t \to 0 $$.
- Also, $$ x = \frac{t}{5} $$.

So the limit becomes
$$
\lim_{t \to 0} \frac{\frac{t}{5}}{\sin t} = \frac{1}{5} \lim_{t \to 0} \frac{t}{\sin t}.
$$

**Step 4.** Recall the standard limit:
$$
\lim_{t\to 0} \frac{t}{\sin t} = 1.
$$
Thus, we have
$$
\lim_{x\to 0^-} x \cot(5x) = \frac{1}{5}.
$$

**Step 5.** Set the left-hand limit equal to $$ f(0) $$ to ensure continuity:
$$
c = \frac{1}{5}.
$$

Hence, the value of $$ c $$ that will make $$ f $$ continuous at $$ x=0 $$ is $$ \frac{1}{5} $$.
To make $$ f $$ continuous at $$ x=0 $$, set $$ c = \frac{1}{5} $$.

5.2. Find the value of that will make continuous at

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