Solve $\lim_{x \rightarrow 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}}$ without using L'Hopital's rule

Question

I tried:

$$\lim_{x \rightarrow 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}} = \frac{e^{x-5}-1}{x-\sqrt{4x+5}} \cdot \frac{x+\sqrt{4x+5}}{x+\sqrt{4x+5}} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-(4x+5)} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-4x-5} = \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x(x-4-\frac{5}{x})} =\\ \frac{(e^{x-5}-1)}{x} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} $$

if I make $y = x-5$ I get

$$ \frac{(e^{y}-1)}{y+5} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = \frac{(e^{y}-1)}{y} \cdot \frac{1}{1+\frac{5}{y}} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = \frac{1}{1+\frac{5}{y}} \cdot \frac{(x+\sqrt{4x+5})}{(x-4-\frac{5}{x})} = ???$$

What do I do next? Am I solving it correctly so far?

Answer 1

Following your calculations, you can proceed: $$\lim_{x\to 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}}=\lim_{x\to 5} \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-4x-5} = \lim_{x\to 5} \frac{e^{x-5}-1}{x-5}\cdot\lim_{x\to 5} \frac{x+\sqrt{4x+5}}{x+1}$$

Now, you can substitute in the first limit $y=x-5$ to get $$\lim_{x\to 5} \frac{e^{x-5}-1}{x-5}=\lim_{y\to 0} \frac{e^y-1}{y}=1$$ since it's the derivative of $e^y$ at $y=0$.

The second limit does not have an indetermined form anymore: $$\lim_{x\to 5} \frac{x+\sqrt{4x+5}}{x+1}=\frac{5+\sqrt{25}}{6}=\frac{10}6=\frac53$$

So in the end, $\lim_{x\to 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}}=\frac 53$.

Answer 2

$$ \begin{aligned} \lim _{x\to 5}\left(\frac{e^{x-5}-1}{x-\sqrt{4x+5}}\right) & = \lim _{t\to 0}\left(\frac{e^{\left(t+5\right)-5}-1}{\left(t+5\right)-\sqrt{4\left(t+5\right)+5}}\right) \\& = \lim _{t\to 0}\left(\frac{e^t-1}{t+5-\sqrt{4t+25}}\right) \\& = \lim _{t\to 0}\left(\frac{\left(1+t+o\left(t\right)\right)-1}{t+5-\left(5+\frac{2}{5}t+o\left(t\right)\right)}\right) \\& = \color{red}{\frac{5}{3}} \end{aligned} $$

Solved with Taylor expansion and substitution of $t = x-5$

Answer 3

I always like to use Taylor developments in those cases. First of all I rename it in terms of $u = x-5$ for comfort

$$ \lim_{u\rightarrow 0}\dfrac{e^u -1}{u+5-\sqrt{4u+25}}$$

The Taylor Development of the numerator term is $$\text{for} \hspace{2mm} u\rightarrow 0, \hspace{6mm}e^u -1 = (1-1) + u + \mathcal{O}(u^2) = u + \mathcal{O}(u^2) $$ and for the denominator:

$$\text{for} \hspace{2mm} u\rightarrow 0, \hspace{5mm} u+5-\sqrt{4u+25} = 5 -\sqrt{25} + u(1 - \dfrac{1}{2}\dfrac{4}{\sqrt{25}}) + \mathcal{O}(u^2) = \dfrac{3}{5} u + \mathcal{O}(u^2) $$

Plugging it back, and ignoring terms that go to 0 faster than linear (quadratics and above):

$$ \lim_{u\rightarrow 0}\dfrac{e^u -1}{u+5-\sqrt{4u+25}} = \lim_{u\rightarrow 0} \dfrac{u + \mathcal{O}(u^2)}{\dfrac{3}{5} u + \mathcal{O}(u^2)} = \dfrac{5}{3}$$

Answer 4

$\lim_{x \rightarrow 5} \frac{e^{x-5}-1}{x-\sqrt{4x+5}}$=$\lim_{x \rightarrow 5} \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{(x-\sqrt{4x+5})(x+\sqrt{4x+5})}$

$\lim_{x \rightarrow 5} \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{x^2-(4x+5)}$ =$\lim_{x \rightarrow 5} \frac{(e^{x-5}-1)(x+\sqrt{4x+5})}{(x+1)(x-5)}$=$\lim_{x \rightarrow 5} \frac{10(e^{x-5}-1))}{6(x-5)}$

Suppose now $x-5=y$ so we have: $\lim_{y \rightarrow 0} \frac{10(e^{y}-1))}{6y}$

And assuming that $e^{y}-1=t$ so we have $y=\ln{(t+1)}$

$\lim_{y \rightarrow 0} \frac{10(e^{y}-1))}{6y}$=$\lim_{t \rightarrow 0} \frac{10}{6\frac{1}{t}(\ln{(t+1))}}$=$\lim_{t \rightarrow 0} \frac{10}{6(\ln{(t+1)^\frac{1}{t})}}$

And also we know that:$\lim_{t \rightarrow 0}({(t+1)^\frac{1}{t})}=e$ so

$\lim_{t \rightarrow 0} \frac{10}{6(\ln{(t+1)^\frac{1}{t})}}=\frac{10}{6}$