Term 213 MaTH 101 3. If \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \) exists, then \( \lim _{z \rightarrow 0} f(x)= \)

To find the value of \( \lim _{x \rightarrow 0} f(x) \) given that \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \) exists, we need to first find the value of \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \) and then use that value to find \( \lim _{x \rightarrow 0} f(x) \). Let's start by finding the value of \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \). Given: \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \) exists To find the value of \( \lim _{x \rightarrow 0} f(x) \), we need to find the value of \( \lim _{x \rightarrow 0} \frac{3 x-2 f(x)+4}{\cos x-e^{x}} \) first. Let's calculate the limit of the given expression as \( x \) approaches 0. Evaluate the limit by following steps: - step0: Evaluate the limit: \(\lim _{x\rightarrow 0}\left(\frac{3x-2fx+4}{\cos\left(x\right)-e^{x}}\right)\) - step1: Evaluate the left-hand and the right-hand limits: \(\begin{align}&\lim _{x\rightarrow 0^{-}}\left(\frac{3x-2fx+4}{\cos\left(x\right)-e^{x}}\right)\\&\lim _{x\rightarrow 0^{+}}\left(\frac{3x-2fx+4}{\cos\left(x\right)-e^{x}}\right)\end{align}\) - step2: Evaluate the left-hand limit: \(\begin{align}&+\infty\\&-\infty\end{align}\) - step3: Limit does not exist: \(\textrm{The limit does not exist}\) The limit of the expression \( \frac{3x-2f(x)+4}{\cos x-e^{x}} \) as \( x \) approaches 0 does not exist. Therefore, the limit of \( f(x) \) as \( x \) approaches 0 does not exist either.