What is the value of the following limit $$\lim_{x\to -1}\frac{2\sqrt{x+5}}{x+1}=?$$ a) $1/2\qquad$ b) $-1/2\qquad$ c) $ 1/4\qquad$ d) $-1/4\qquad$
my try: i let $x+1=t$, so i get
$$\lim_{x\to -1}\frac{2\sqrt{x+5}}{x+1}=\lim_{t \to 0}\frac{2\sqrt{t+4}}{t}$$ $$=2\lim_{t \to 0}\frac{2\sqrt{1+\frac{t}{4}}}{t}$$ $$=4\lim_{t \to 0}\frac{\left(1+\frac{t}{4}\right)^{1/2}}{t}$$ then using binomial expansion for $|x|<1$, $(1+x)^n=1+nx+\frac{n(n-1)}{2}x^2+\ldots$, i get $$4\lim_{t \to 0}\frac{1+\frac12\frac{t}{4}+\frac{(1/2)(1/2-1)}{2!}\left(\frac{t}{4}\right)^2+\ldots}{t}$$ $$=4\lim_{t \to 0}(\frac1t+\frac{1}{8}-\frac{t}{128}+\ldots)$$ here limit tends to infinity but my book says the limit is $1/2$, I don't know how. please help me solve this limit. suggest me other method if any. thanks
The written limit does not exist, as it goes to $-\infty$ from the left and to $\infty$ from the right, because the numerator goes to $4$ and the denominator to $0$.
There is something missing: $$ \lim_{x\to-1}\frac{2\sqrt{x+5}-4}{x+1} $$ is the most probable candidate to be the correct limit to compute.