$$\lim_{t \to 0} \frac{4}{t\sqrt{16+t}} {-} {\frac1t}$$

i have tried many times but i could get the correct answer

$$\lim_{t \to 0}\frac{4-\sqrt{16+t}}{t\sqrt{16+t}}$$

$$\lim_{t \to 0}\left(\frac{4-\sqrt{16+t}}{t\sqrt{16+t}} \cdot {\frac{4+\sqrt{16+t}}{4+\sqrt{16+t}}}\right)$$ $$\lim_{t \to 0}\left(\frac{16-(16+t)}{4t\sqrt{16+t}+(t(16+t))} \right)$$ $$\lim_{t \to 0}\left(\frac{-1}{4\sqrt{16+t}+(16+t)} \right)$$ the above step is my calculation but i cannot figure out what is wrong with it.

any help will be appreciate.

  • 1
    $\begingroup$ You are almost there! Just let $t$ go to zero in the last expression. What do you get? $\endgroup$ – bames Feb 14 '18 at 9:04
  • $\begingroup$ oh i think i have messed up with the +ve and -ve sign $\endgroup$ – fds Feb 14 '18 at 9:20

Nothing is wrong !

$\lim_{t \to 0}\left(\frac{-1}{4\sqrt{16+t}+(16+t)} \right)=- \frac{1}{4 \sqrt{16}+16}= ....$.

  • $\begingroup$ thanks for helping $\endgroup$ – fds Feb 14 '18 at 9:21

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