Maths (commerce) Solve the Following Question.(2 Marks)

Evaluate the limit of the function if exist at $x=1$ where,
$
f(x)= \begin{cases}7-4 x & x<1 \\ x^2+2 & x \geq 1\end{cases}
$

$\lim _{x \rightarrow 0}\left[\frac{\log 100+\log (0.01+x)}{x}\right]$

$\lim _{x \rightarrow 0}\left[\frac{a^{4 x}-1}{b^{2 x}-1}\right]$

$\lim _{x \rightarrow 0}\left[\frac{\left(5^x-1\right)^2}{x \cdot \log (1+x)}\right]$

$\lim _{x \rightarrow 0}\left[\frac{a^{3 x}-a^{2 x}-a^x+1}{x^2}\right]$

$\lim _{x \rightarrow 0} \frac{e^x+e^{-x}-2}{x^2}$

$\lim _{x \rightarrow 0}\left[\frac{a^x+b^x+c^x-3}{x}\right]$

$\lim _{x \rightarrow a} \frac{(x+2)^{\frac{5}{3}}-(a+2)^{\frac{5}{3}}}{x-a}$

$\lim _{x \rightarrow 0}\left[\frac{(25)^x-2(5)^x+1}{x^2}\right]$

$\lim _{x \rightarrow 0}\left[\frac{15^x-5^x-3^x+1}{x^2}\right]$

$\lim _{x \rightarrow 0}\left[\frac{5^x+3^x-2^x-1}{x}\right]$

$\lim _{x \rightarrow 0}\left[\frac{9^x-5^x}{4^x-1}\right]$

$\lim _{y \rightarrow 2}\left[\frac{2-y}{\sqrt{3-y}-1}\right]$

$\lim _{x \rightarrow 1}\left[\frac{x^2+x \sqrt{x}-2}{x-1}\right]$

$\lim _{y \rightarrow 0}\left[\frac{\sqrt{1-y^2}-\sqrt{1+y^2}}{y^2}\right]$

$\lim _{x \rightarrow 0}\left[\frac{\sqrt{6+x+x^2}-\sqrt{6}}{x}\right]$

$\lim _{x \rightarrow 3}\left[\frac{1}{x-3}-\frac{9 x}{x^3-27}\right]$

$\lim _{x \rightarrow 5}\left[\frac{x^3-125}{x^2-25}\right]$

$\lim _{z \rightarrow a}\left[\frac{(z+2)^{\frac{3}{2}}-(a+2)^{\frac{3}{2}}}{z-a}\right]$

If $\lim _{x \rightarrow 5}\left[\frac{x^k-5^k}{x-5}\right]=500$, find all possible values of $\mathrm{k}$.

$\lim _{x \rightarrow 7}\left[\frac{(\sqrt[3]{x}-\sqrt[3]{7})(\sqrt[3]{x}+\sqrt[3]{7})}{x-7}\right]$

$\lim _{x \rightarrow 5}\left[\frac{x^3-125}{x^5-3125}\right]$