Question

Find the $\underset{x\to 0}{lim}\frac{{(x+4)}^{3/2}+e^x-9}{x}$ without using L'Hospital rule

Answer 1

$(x+4{)}^{3/2}=8(1+\frac{x}{4}{)}^{3/2}$

$(1+x/m{)}^n=1+n\frac{x}{m}+\frac{n(n-1)}{2!}\frac{x^2}{m^2}+.....$

Hence expansion of $(1+\frac{x}{4}{)}^{3/2}$ $=1+\frac{3}{2}x$ $+\frac{(\frac{3}{2})(\frac{1}{2})}{2!}{x}^2.....$

now expansion of $e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+.....$

now put these expansion on given expression

$\underset{x\to 0}{lim}\frac{(x+4{)}^{3/2}+e^x-9}{x}$ $=\frac{8(1+\frac{3}{2}\frac{x}{4}+\frac{3}{8}\frac{x^2}{4^2})+(1+\frac{x}{1!}+\frac{x^2}{2!})-9}{x}$

On solving this equation we will get

$\underset{x\to 0}{lim}\frac{(x+4{)}^{3/2}+e^x-9}{x}=4$